diff --git a/Python/balanced-binary-tree.py b/Python/balanced-binary-tree.py index fbf1326ab..c65c301ba 100644 --- a/Python/balanced-binary-tree.py +++ b/Python/balanced-binary-tree.py @@ -3,37 +3,32 @@ # # Given a binary tree, determine if it is height-balanced. # -# For this problem, a height-balanced binary tree is defined as a binary tree -# in which the depth of the two subtrees of every node never differ by more than 1. +# For this problem, a height-balanced binary tree is defined as a binary +# tree +# in which the depth of the two subtrees of every node never differ by more +# than 1. # + # Definition for a binary tree node -class TreeNode: +class TreeNode(object): def __init__(self, x): self.val = x self.left = None self.right = None -class Solution: + +class Solution(object): # @param root, a tree node # @return a boolean def isBalanced(self, root): - return (self.getHeight(root) >= 0) - - def getHeight(self, root): - if root is None: - return 0 - left_height, right_height = self.getHeight(root.left), self.getHeight(root.right) - if left_height < 0 or right_height < 0 or abs(left_height - right_height) > 1: - return -1 - return max(left_height, right_height) + 1 - -if __name__ == "__main__": - root = TreeNode(0) - root.left = TreeNode(1) - result = Solution().isBalanced(root) - print result - - root.left.left = TreeNode(2) - result = Solution().isBalanced(root) - print result + def getHeight(root): + if root is None: + return 0 + left_height, right_height = \ + getHeight(root.left), getHeight(root.right) + if left_height < 0 or right_height < 0 or \ + abs(left_height - right_height) > 1: + return -1 + return max(left_height, right_height) + 1 + return (getHeight(root) >= 0) diff --git a/Python/base-7.py b/Python/base-7.py index b6544bd02..5fbc956a7 100644 --- a/Python/base-7.py +++ b/Python/base-7.py @@ -11,9 +11,11 @@ # Output: "-10" # Note: The input will be in range of [-1e7, 1e7]. + class Solution(object): def convertToBase7(self, num): - if num < 0: return '-' + self.convertToBase7(-num) + if num < 0: + return '-' + self.convertToBase7(-num) result = '' while num: result = str(num % 7) + result @@ -27,6 +29,8 @@ def convertToBase7(self, num): :type num: int :rtype: str """ - if num < 0: return '-' + self.convertToBase7(-num) - if num < 7: return str(num) + if num < 0: + return '-' + self.convertToBase7(-num) + if num < 7: + return str(num) return self.convertToBase7(num // 7) + str(num % 7) diff --git a/Python/baseball-game.py b/Python/baseball-game.py index ede8ff9d9..309e1a45f 100644 --- a/Python/baseball-game.py +++ b/Python/baseball-game.py @@ -4,38 +4,45 @@ # You're now a baseball game point recorder. # Given a list of strings, each string can be one of the 4 following types: # -# 1. Integer (one round's score): Directly represents the number of points you get in this round. -# 2. "+" (one round's score): Represents that the points you get in this round are -# the sum of the last two valid round's points. -# 3. "D" (one round's score): Represents that the points you get in this round are -# the doubled data of the last valid round's points. -# 4. "C" (an operation, which isn't a round's score): Represents the last valid round's points -# you get were invalid and should be removed. -# -# Each round's operation is permanent and could have an impact on the round before and the round after. +# 1. Integer (one round's score): Directly represents the number of points +# you get in this round. +# 2. "+" (one round's score): Represents that the points you get in +# this round are the sum of the last two valid +# round's points. +# 3. "D" (one round's score): Represents that the points you get in this round +# are the doubled data of the last valid round's +# points. +# 4. "C" (an operation, which isn't a round's score): Represents the last +# valid round's points you get were invalid and +# should be removed. +# +# Each round's operation is permanent and could have an impact on the round +# before and the round after. # You need to return the sum of the points you could get in all the rounds. # # Example 1: # # Input: ["5","2","C","D","+"] # Output: 30 -# Explanation: +# Explanation: # Round 1: You could get 5 points. The sum is: 5. # Round 2: You could get 2 points. The sum is: 7. -# Operation 1: The round 2's data was invalid. The sum is: 5. -# Round 3: You could get 10 points (the round 2's data has been removed). The sum is: 15. +# Operation 1: The round 2's data was invalid. The sum is: 5. +# Round 3: You could get 10 points (the round 2's data has been removed). +# The sum is: 15. # Round 4: You could get 5 + 10 = 15 points. The sum is: 30. -# +# # Example 2: # # Input: ["5","-2","4","C","D","9","+","+"] # Output: 27 -# Explanation: +# Explanation: # Round 1: You could get 5 points. The sum is: 5. # Round 2: You could get -2 points. The sum is: 3. # Round 3: You could get 4 points. The sum is: 7. -# Operation 1: The round 3's data is invalid. The sum is: 3. -# Round 4: You could get -4 points (the round 3's data has been removed). The sum is: -1. +# Operation 1: The round 3's data is invalid. The sum is: 3. +# Round 4: You could get -4 points (the round 3's data has been removed). +# The sum is: -1. # Round 5: You could get 9 points. The sum is: 8. # Round 6: You could get -4 + 9 = 5 points. The sum is 13. # Round 7: You could get 9 + 5 = 14 points. The sum is 27. @@ -43,7 +50,8 @@ # Note: # The size of the input list will be between 1 and 1000. # Every integer represented in the list will be between -30000 and 30000. - + + class Solution(object): def calPoints(self, ops): """ diff --git a/Python/basic-calculator-ii.py b/Python/basic-calculator-ii.py index 4575b8a69..03df2953e 100644 --- a/Python/basic-calculator-ii.py +++ b/Python/basic-calculator-ii.py @@ -3,7 +3,7 @@ # # Implement a basic calculator to evaluate a simple expression string. # -# The expression string contains only non-negative integers, +, -, *, / +# The expression string contains only non-negative integers, +, -, *, / # operators and empty spaces . The integer division should truncate toward zero. # # You may assume that the given expression is always valid. @@ -38,10 +38,10 @@ def calculate(self, s): while operators[-1] != ')': self.compute(operands, operators) operators.pop() - + while operators: self.compute(operands, operators) - + return operands[-1] def compute(self, operands, operators): diff --git a/Python/basic-calculator-iii.py b/Python/basic-calculator-iii.py index 0010d2a4d..04568b2cf 100644 --- a/Python/basic-calculator-iii.py +++ b/Python/basic-calculator-iii.py @@ -46,10 +46,10 @@ def calculate(self, s): while operators[-1] != ')': self.compute(operands, operators) operators.pop() - + while operators: self.compute(operands, operators) - + return operands[-1] def compute(self, operands, operators): @@ -63,4 +63,4 @@ def compute(self, operands, operators): operands.append(left * right) elif op == '/': operands.append(left / right) - + diff --git a/Python/basic-calculator.py b/Python/basic-calculator.py index ea4ca3245..cae7f5527 100644 --- a/Python/basic-calculator.py +++ b/Python/basic-calculator.py @@ -32,10 +32,10 @@ def calculate(self, s): while operators[-1] != ')': self.compute(operands, operators) operators.pop() - + while operators: self.compute(operands, operators) - + return operands[-1] def compute(self, operands, operators): diff --git a/Python/battleships-in-a-board.py b/Python/battleships-in-a-board.py index 9f9c49944..7d6c7490b 100644 --- a/Python/battleships-in-a-board.py +++ b/Python/battleships-in-a-board.py @@ -9,7 +9,7 @@ # Battleships can only be placed horizontally or vertically. In other words, # they can only be made of the shape 1xN (1 row, N columns) or Nx1 (N rows, 1 column), # where N can be of any size. -# At least one horizontal or vertical cell separates between two battleships - +# At least one horizontal or vertical cell separates between two battleships - # there are no adjacent battleships. # # Example: diff --git a/Python/beautiful-arrangement-ii.py b/Python/beautiful-arrangement-ii.py index 007a28eb6..e9617beec 100644 --- a/Python/beautiful-arrangement-ii.py +++ b/Python/beautiful-arrangement-ii.py @@ -3,7 +3,7 @@ # Given two integers n and k, # you need to construct a list which contains n different positive integers ranging -# from 1 to n and obeys the following requirement: +# from 1 to n and obeys the following requirement: # Suppose this list is [a1, a2, a3, ... , an], # then the list [|a1 - a2|, |a2 - a3|, |a3 - a4|, ... , |an-1 - an|] has exactly k distinct integers. # diff --git a/Python/beautiful-arrangement.py b/Python/beautiful-arrangement.py index 383209c20..64327bfcd 100644 --- a/Python/beautiful-arrangement.py +++ b/Python/beautiful-arrangement.py @@ -2,7 +2,7 @@ # Space: O(n) # Suppose you have N integers from 1 to N. -# We define a beautiful arrangement as an array that is constructed by these N numbers successfully +# We define a beautiful arrangement as an array that is constructed by these N numbers successfully # if one of the following is true for the ith position (1 <= i <= N) in this array: # # The number at the ith position is divisible by i. @@ -12,7 +12,7 @@ # Example 1: # Input: 2 # Output: 2 -# Explanation: +# Explanation: # # The first beautiful arrangement is [1, 2]: # diff --git a/Python/best-meeting-point.py b/Python/best-meeting-point.py index 52b481624..ca5554f63 100644 --- a/Python/best-meeting-point.py +++ b/Python/best-meeting-point.py @@ -13,10 +13,10 @@ def minTotalDistance(self, grid): y = [j for row in grid for j, v in enumerate(row) if v == 1] mid_x = self.findKthLargest(x, len(x) / 2 + 1) mid_y = self.findKthLargest(y, len(y) / 2 + 1) - + return sum([abs(mid_x-i) + abs(mid_y-j) \ for i, row in enumerate(grid) for j, v in enumerate(row) if v == 1]) - + def findKthLargest(self, nums, k): left, right = 0, len(nums) - 1 while left <= right: @@ -28,7 +28,7 @@ def findKthLargest(self, nums, k): right = new_pivot_idx - 1 else: # new_pivot_idx < k - 1. left = new_pivot_idx + 1 - + def PartitionAroundPivot(self, left, right, pivot_idx, nums): pivot_value = nums[pivot_idx] new_pivot_idx = left @@ -37,6 +37,6 @@ def PartitionAroundPivot(self, left, right, pivot_idx, nums): if nums[i] > pivot_value: nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i] new_pivot_idx += 1 - + nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right] return new_pivot_idx diff --git a/Python/best-time-to-buy-and-sell-stock-ii.py b/Python/best-time-to-buy-and-sell-stock-ii.py index 133ed078a..b0d12c7a9 100644 --- a/Python/best-time-to-buy-and-sell-stock-ii.py +++ b/Python/best-time-to-buy-and-sell-stock-ii.py @@ -1,13 +1,13 @@ # Time: O(n) # Space: O(1) # -# Say you have an array for which the ith element is +# Say you have an array for which the ith element is # the price of a given stock on day i. # -# Design an algorithm to find the maximum profit. -# You may complete as many transactions as you like -# (ie, buy one and sell one share of the stock multiple times). -# However, you may not engage in multiple transactions at the same time +# Design an algorithm to find the maximum profit. +# You may complete as many transactions as you like +# (ie, buy one and sell one share of the stock multiple times). +# However, you may not engage in multiple transactions at the same time # (ie, you must sell the stock before you buy again). @@ -17,7 +17,7 @@ class Solution: def maxProfit(self, prices): profit = 0 for i in xrange(len(prices) - 1): - profit += max(0, prices[i + 1] - prices[i]) + profit += max(0, prices[i + 1] - prices[i]) return profit def maxProfit2(self, prices): diff --git a/Python/best-time-to-buy-and-sell-stock-iii.py b/Python/best-time-to-buy-and-sell-stock-iii.py index 114c1b321..89109f853 100644 --- a/Python/best-time-to-buy-and-sell-stock-iii.py +++ b/Python/best-time-to-buy-and-sell-stock-iii.py @@ -1,14 +1,14 @@ # Time: O(n) # Space: O(1) # -# Say you have an array for which the ith element +# Say you have an array for which the ith element # is the price of a given stock on day i. # -# Design an algorithm to find the maximum profit. +# Design an algorithm to find the maximum profit. # You may complete at most two transactions. # # Note: -# You may not engage in multiple transactions at the same time +# You may not engage in multiple transactions at the same time # (ie, you must sell the stock before you buy again). # @@ -26,7 +26,7 @@ def maxProfit(self, prices): release1 = max(release1, hold1 + i) hold1 = max(hold1, -i); return release2 - + # Time: O(k * n) # Space: O(k) class Solution2: @@ -34,7 +34,7 @@ class Solution2: # @return an integer def maxProfit(self, prices): return self.maxAtMostKPairsProfit(prices, 2) - + def maxAtMostKPairsProfit(self, prices, k): max_buy = [float("-inf") for _ in xrange(k + 1)] max_sell = [0 for _ in xrange(k + 1)] @@ -47,7 +47,7 @@ def maxAtMostKPairsProfit(self, prices, k): return max_sell[k] # Time: O(n) -# Space: O(n) +# Space: O(n) class Solution3: # @param prices, a list of integer # @return an integer @@ -57,19 +57,19 @@ def maxProfit(self, prices): min_price = min(min_price, price) max_profit_from_left = max(max_profit_from_left, price - min_price) max_profits_from_left.append(max_profit_from_left) - + max_price, max_profit_from_right, max_profits_from_right = 0, 0, [] for i in reversed(range(len(prices))): max_price = max(max_price, prices[i]) max_profit_from_right = max(max_profit_from_right, max_price - prices[i]) max_profits_from_right.insert(0, max_profit_from_right) - + max_profit = 0 for i in range(len(prices)): max_profit = max(max_profit, max_profits_from_left[i] + max_profits_from_right[i]) - + return max_profit - + if __name__ == "__main__": result = Solution().maxProfit([3, 2, 1, 4, 2, 5, 6]) print result diff --git a/Python/best-time-to-buy-and-sell-stock-iv.py b/Python/best-time-to-buy-and-sell-stock-iv.py index 0235f07df..2d0071909 100644 --- a/Python/best-time-to-buy-and-sell-stock-iv.py +++ b/Python/best-time-to-buy-and-sell-stock-iv.py @@ -2,15 +2,15 @@ # Space: O(k) # # Say you have an array for which the ith element is the price of a given stock on day i. -# +# # Design an algorithm to find the maximum profit. You may complete at most k transactions. -# +# # Note: # You may not engage in multiple transactions at the same time (ie, you must sell the stock before you buy again). # class Solution: - # @return an integer as the maximum profit + # @return an integer as the maximum profit def maxProfit(self, k, prices): if k >= len(prices) / 2: return self.maxAtMostNPairsProfit(prices) @@ -20,7 +20,7 @@ def maxProfit(self, k, prices): def maxAtMostNPairsProfit(self, prices): profit = 0 for i in xrange(len(prices) - 1): - profit += max(0, prices[i + 1] - prices[i]) + profit += max(0, prices[i + 1] - prices[i]) return profit def maxAtMostKPairsProfit(self, prices, k): diff --git a/Python/best-time-to-buy-and-sell-stock-with-transaction-fee.py b/Python/best-time-to-buy-and-sell-stock-with-transaction-fee.py index 7393b5b6e..4b622e833 100644 --- a/Python/best-time-to-buy-and-sell-stock-with-transaction-fee.py +++ b/Python/best-time-to-buy-and-sell-stock-with-transaction-fee.py @@ -39,4 +39,4 @@ def maxProfit(self, prices, fee): cash = max(cash, hold+prices[i]-fee) hold = max(hold, cash-prices[i]) return cash - + diff --git a/Python/best-time-to-buy-and-sell-stock.py b/Python/best-time-to-buy-and-sell-stock.py index 2c86c4a74..748b5e638 100644 --- a/Python/best-time-to-buy-and-sell-stock.py +++ b/Python/best-time-to-buy-and-sell-stock.py @@ -1,11 +1,11 @@ # Time: O(n) # Space: O(1) # -# Say you have an array for which the ith element +# Say you have an array for which the ith element # is the price of a given stock on day i. # -# If you were only permitted to complete at most one transaction -# (ie, buy one and sell one share of the stock), +# If you were only permitted to complete at most one transaction +# (ie, buy one and sell one share of the stock), # design an algorithm to find the maximum profit. # @@ -16,10 +16,10 @@ def maxProfit(self, prices): max_profit, min_price = 0, float("inf") for price in prices: min_price = min(min_price, price) - max_profit = max(max_profit, price - min_price) + max_profit = max(max_profit, price - min_price) return max_profit if __name__ == "__main__": result = Solution().maxProfit([3, 2, 1, 4, 2, 5, 6]) print result - + diff --git a/Python/binary-number-with-alternating-bits.py b/Python/binary-number-with-alternating-bits.py index ff066353a..e8f06d5b9 100644 --- a/Python/binary-number-with-alternating-bits.py +++ b/Python/binary-number-with-alternating-bits.py @@ -27,7 +27,7 @@ # Output: True # Explanation: # The binary representation of 10 is: 1010. - + class Solution(object): def hasAlternatingBits(self, n): """ diff --git a/Python/binary-search-tree-iterator.py b/Python/binary-search-tree-iterator.py index 428c8d4a3..ae455dea3 100644 --- a/Python/binary-search-tree-iterator.py +++ b/Python/binary-search-tree-iterator.py @@ -1,11 +1,11 @@ # Time: O(1) # Space: O(h), h is height of binary tree -# +# # Implement an iterator over a binary search tree (BST). # Your iterator will be initialized with the root node of a BST. -# +# # Calling next() will return the next smallest number in the BST. -# +# # Note: next() and hasNext() should run in average O(1) time # and uses O(h) memory, where h is the height of the tree. # @@ -32,19 +32,19 @@ def next(self): while self.cur: self.stack.append(self.cur) self.cur = self.cur.left - + self.cur = self.stack.pop() node = self.cur self.cur = self.cur.right - + return node.val - + if __name__ == "__main__": root = TreeNode(2) root.left = TreeNode(1) - + # Your BSTIterator will be called like this: i, v = BSTIterator(root), [] while i.hasNext(): v.append(i.next()) - + print v diff --git a/Python/binary-tree-inorder-traversal.py b/Python/binary-tree-inorder-traversal.py index 11448e3fa..b82319cfe 100644 --- a/Python/binary-tree-inorder-traversal.py +++ b/Python/binary-tree-inorder-traversal.py @@ -39,7 +39,7 @@ def inorderTraversal(self, root): node = curr.left while node.right and node.right != curr: node = node.right - + if node.right is None: node.right = curr curr = curr.left @@ -47,7 +47,7 @@ def inorderTraversal(self, root): result.append(curr.val) node.right = None curr = curr.right - + return result diff --git a/Python/binary-tree-level-order-traversal-ii.py b/Python/binary-tree-level-order-traversal-ii.py index a03432a5d..0973c82e2 100644 --- a/Python/binary-tree-level-order-traversal-ii.py +++ b/Python/binary-tree-level-order-traversal-ii.py @@ -3,7 +3,7 @@ # Given a binary tree, return the bottom-up level order traversal of its nodes' values. # (ie, from left to right, level by level from leaf to root). -# +# # For example: # Given binary tree {3,9,20,#,#,15,7}, # 3 diff --git a/Python/binary-tree-level-order-traversal.py b/Python/binary-tree-level-order-traversal.py index f1b8c9c47..8e48c8372 100644 --- a/Python/binary-tree-level-order-traversal.py +++ b/Python/binary-tree-level-order-traversal.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(n) # -# Given a binary tree, return the level order traversal of its nodes' values. +# Given a binary tree, return the level order traversal of its nodes' values. # (ie, from left to right, level by level). # # For example: diff --git a/Python/binary-tree-longest-consecutive-sequence-ii.py b/Python/binary-tree-longest-consecutive-sequence-ii.py index 99c88f13c..b7c0f7bec 100644 --- a/Python/binary-tree-longest-consecutive-sequence-ii.py +++ b/Python/binary-tree-longest-consecutive-sequence-ii.py @@ -36,4 +36,4 @@ def longestConsecutiveHelper(root): self.max_len = 0 longestConsecutiveHelper(root) return self.max_len - + diff --git a/Python/binary-tree-longest-consecutive-sequence.py b/Python/binary-tree-longest-consecutive-sequence.py index 769c417ee..e9c6c7444 100644 --- a/Python/binary-tree-longest-consecutive-sequence.py +++ b/Python/binary-tree-longest-consecutive-sequence.py @@ -19,10 +19,10 @@ def longestConsecutive(self, root): def longestConsecutiveHelper(root): if not root: return 0 - + left_len = longestConsecutiveHelper(root.left) right_len = longestConsecutiveHelper(root.right) - + cur_len = 1 if root.left and root.left.val == root.val + 1: cur_len = max(cur_len, left_len + 1); diff --git a/Python/binary-tree-maximum-path-sum.py b/Python/binary-tree-maximum-path-sum.py index d71f6eeba..b3d25e94a 100644 --- a/Python/binary-tree-maximum-path-sum.py +++ b/Python/binary-tree-maximum-path-sum.py @@ -2,12 +2,12 @@ # Space: O(h), h is height of binary tree # # Given a binary tree, find the maximum path sum. -# +# # The path may start and end at any node in the tree. -# +# # For example: # Given the below binary tree, -# +# # 1 # / \ # 2 3 @@ -27,7 +27,7 @@ class Solution: def maxPathSum(self, root): self.maxPathSumRecu(root) return self.maxSum - + def maxPathSumRecu(self, root): if root is None: return 0 @@ -35,7 +35,7 @@ def maxPathSumRecu(self, root): right = max(0, self.maxPathSumRecu(root.right)) self.maxSum = max(self.maxSum, root.val + left + right) return root.val + max(left, right) - + if __name__ == "__main__": root = TreeNode(1) root.left = TreeNode(2) diff --git a/Python/binary-tree-paths.py b/Python/binary-tree-paths.py index 657e8d579..770adfdfb 100644 --- a/Python/binary-tree-paths.py +++ b/Python/binary-tree-paths.py @@ -29,7 +29,7 @@ def binaryTreePaths(self, root): result, path = [], [] self.binaryTreePathsRecu(root, path, result) return result - + def binaryTreePathsRecu(self, node, path, result): if node is None: return diff --git a/Python/binary-tree-postorder-traversal.py b/Python/binary-tree-postorder-traversal.py index c5a47b0c8..e0ae110cd 100644 --- a/Python/binary-tree-postorder-traversal.py +++ b/Python/binary-tree-postorder-traversal.py @@ -40,7 +40,7 @@ def postorderTraversal(self, root): node = cur.left while node.right and node.right != cur: node = node.right - + if node.right is None: node.right = cur cur = cur.left @@ -48,9 +48,9 @@ def postorderTraversal(self, root): result += self.traceBack(cur.left, node) node.right = None cur = cur.right - + return result - + def traceBack(self, frm, to): result, cur = [], frm while cur is not to: @@ -63,7 +63,7 @@ def traceBack(self, frm, to): # Time: O(n) # Space: O(h) -# Stack Solution +# Stack Solution class Solution2(object): def postorderTraversal(self, root): """ diff --git a/Python/binary-tree-preorder-traversal.py b/Python/binary-tree-preorder-traversal.py index 2ba22d545..9cf12ee7a 100644 --- a/Python/binary-tree-preorder-traversal.py +++ b/Python/binary-tree-preorder-traversal.py @@ -39,7 +39,7 @@ def preorderTraversal(self, root): node = curr.left while node.right and node.right != curr: node = node.right - + if node.right is None: result.append(curr.val) node.right = curr @@ -47,13 +47,13 @@ def preorderTraversal(self, root): else: node.right = None curr = curr.right - + return result # Time: O(n) # Space: O(h) -# Stack Solution +# Stack Solution class Solution2(object): def preorderTraversal(self, root): """ diff --git a/Python/binary-tree-right-side-view.py b/Python/binary-tree-right-side-view.py index 56d5e3db2..effd0ad13 100644 --- a/Python/binary-tree-right-side-view.py +++ b/Python/binary-tree-right-side-view.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(h) # -# Given a binary tree, imagine yourself standing on the right side of it, +# Given a binary tree, imagine yourself standing on the right side of it, # return the values of the nodes you can see ordered from top to bottom. # # For example: @@ -28,27 +28,27 @@ def rightSideView(self, root): result = [] self.rightSideViewDFS(root, 1, result) return result - + def rightSideViewDFS(self, node, depth, result): if not node: return - + if depth > len(result): result.append(node.val) - + self.rightSideViewDFS(node.right, depth+1, result) self.rightSideViewDFS(node.left, depth+1, result) # BFS solution # Time: O(n) -# Space: O(n) +# Space: O(n) class Solution2: # @param root, a tree node # @return a list of integers def rightSideView(self, root): if root is None: return [] - + result, current = [], [root] while current: next_level = [] @@ -60,7 +60,7 @@ def rightSideView(self, root): if i == len(current) - 1: result.append(node.val) current = next_level - + return result if __name__ == "__main__": diff --git a/Python/binary-tree-tilt.py b/Python/binary-tree-tilt.py index e1ed02783..76ca7e745 100644 --- a/Python/binary-tree-tilt.py +++ b/Python/binary-tree-tilt.py @@ -10,12 +10,12 @@ # The tilt of the whole tree is defined as the sum of all nodes' tilt. # # Example: -# Input: +# Input: # 1 # / \ # 2 3 # Output: 1 -# Explanation: +# Explanation: # Tilt of node 2 : 0 # Tilt of node 3 : 0 # Tilt of node 1 : |2-3| = 1 diff --git a/Python/binary-tree-upside-down.py b/Python/binary-tree-upside-down.py index 51315ff94..e68f4e923 100644 --- a/Python/binary-tree-upside-down.py +++ b/Python/binary-tree-upside-down.py @@ -1,27 +1,27 @@ # Time: O(n) # Space: O(1) # -# Given a binary tree where all the right nodes are either leaf nodes with a sibling -# (a left node that shares the same parent node) or empty, flip it upside down and -# turn it into a tree where the original right nodes turned into left leaf nodes. +# Given a binary tree where all the right nodes are either leaf nodes with a sibling +# (a left node that shares the same parent node) or empty, flip it upside down and +# turn it into a tree where the original right nodes turned into left leaf nodes. # Return the new root. -# +# # For example: # Given a binary tree {1,2,3,4,5}, -# +# # 1 # / \ # 2 3 # / \ # 4 5 -# +# # return the root of the binary tree [4,5,2,#,#,3,1]. -# +# # 4 # / \ # 5 2 # / \ -# 3 1 +# 3 1 # # Definition for a binary tree node @@ -36,7 +36,7 @@ class Solution: # @return root of the upside down tree def upsideDownBinaryTree(self, root): p, parent, parent_right = root, None, None - + while p: left = p.left p.left = parent_right @@ -44,7 +44,7 @@ def upsideDownBinaryTree(self, root): p.right = parent parent = p p = left - + return parent # Time: O(n) @@ -54,16 +54,16 @@ class Solution2: # @return root of the upside down tree def upsideDownBinaryTree(self, root): return self.upsideDownBinaryTreeRecu(root, None) - + def upsideDownBinaryTreeRecu(self, p, parent): if p is None: return parent - + root = self.upsideDownBinaryTreeRecu(p.left, p) if parent: p.left = parent.right else: p.left = None p.right = parent - + return root \ No newline at end of file diff --git a/Python/binary-tree-zigzag-level-order-traversal.py b/Python/binary-tree-zigzag-level-order-traversal.py index 7a6261e05..1a5dac583 100644 --- a/Python/binary-tree-zigzag-level-order-traversal.py +++ b/Python/binary-tree-zigzag-level-order-traversal.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between). -# +# # For example: # Given binary tree {3,9,20,#,#,15,7}, # 3 diff --git a/Python/bitwise-and-of-numbers-range.py b/Python/bitwise-and-of-numbers-range.py index 369a5a803..0dc2cad71 100644 --- a/Python/bitwise-and-of-numbers-range.py +++ b/Python/bitwise-and-of-numbers-range.py @@ -1,7 +1,7 @@ # Time: O(1) # Space: O(1) # -# Given a range [m, n] where 0 <= m <= n <= 2147483647, +# Given a range [m, n] where 0 <= m <= n <= 2147483647, # return the bitwise AND of all numbers in this range, inclusive. # # For example, given the range [5, 7], you should return 4. diff --git a/Python/boundary-of-binary-tree.py b/Python/boundary-of-binary-tree.py index ad7a6a6b3..c71ed0df4 100644 --- a/Python/boundary-of-binary-tree.py +++ b/Python/boundary-of-binary-tree.py @@ -31,7 +31,7 @@ def rightBoundary(root, nodes): else: rightBoundary(root.right, nodes) nodes.append(root.val) - + def leaves(root, nodes): if not root: return @@ -43,7 +43,7 @@ def leaves(root, nodes): if not root: return [] - + nodes = [root.val] leftBoundary(root.left, nodes) leaves(root.left, nodes) diff --git a/Python/bricks-falling-when-hit.py b/Python/bricks-falling-when-hit.py index 3a94f4bd5..9901572c7 100644 --- a/Python/bricks-falling-when-hit.py +++ b/Python/bricks-falling-when-hit.py @@ -12,19 +12,19 @@ # Return an array representing the number of bricks that will drop after each erasure in sequence. # # Example 1: -# Input: +# Input: # grid = [[1,0,0,0],[1,1,1,0]] # hits = [[1,0]] # Output: [2] -# Explanation: +# Explanation: # If we erase the brick at (1, 0), the brick at (1, 1) and (1, 2) will drop. So we should return 2. # # Example 2: -# Input: +# Input: # grid = [[1,0,0,0],[1,1,0,0]] # hits = [[1,1],[1,0]] # Output: [0,0] -# Explanation: +# Explanation: # When we erase the brick at (1, 0), the brick at (1, 1) has already disappeared due to the last move. # So each erasure will cause no bricks dropping. # Note that the erased brick (1, 0) will not be counted as a dropped brick. @@ -56,7 +56,7 @@ def union_set(self, x, y): def top(self): return self.size[self.find_set(len(self.size)-1)] - + class Solution(object): def hitBricks(self, grid, hits): @@ -67,7 +67,7 @@ def hitBricks(self, grid, hits): """ def index(C, r, c): return r*C+c - + directions = [(0, -1), (0, 1), (-1, 0), (1, 0)] R, C = len(grid), len(grid[0]) diff --git a/Python/bulb-switcher-ii.py b/Python/bulb-switcher-ii.py index 3ce93572a..3bb37192d 100644 --- a/Python/bulb-switcher-ii.py +++ b/Python/bulb-switcher-ii.py @@ -24,7 +24,7 @@ # Output: 4 # Explanation: Status can be: [off, on, off], [on, off, on], [off, off, off], [off, on, on]. # Note: n and m both fit in range [0, 1000]. - + class Solution(object): def flipLights(self, n, m): """ diff --git a/Python/burst-balloons.py b/Python/burst-balloons.py index acdd5c15e..4cb6ebd9c 100644 --- a/Python/burst-balloons.py +++ b/Python/burst-balloons.py @@ -1,20 +1,20 @@ # Time: O(n^3) # Space: O(n^2) -# Given n balloons, indexed from 0 to n-1. -# Each balloon is painted with a number on it -# represented by array nums. +# Given n balloons, indexed from 0 to n-1. +# Each balloon is painted with a number on it +# represented by array nums. # You are asked to burst all the balloons. -# If the you burst balloon i you will get -# nums[left] * nums[i] * nums[right] coins. -# Here left and right are adjacent indices of i. +# If the you burst balloon i you will get +# nums[left] * nums[i] * nums[right] coins. +# Here left and right are adjacent indices of i. # After the burst, the left and right then # becomes adjacent. # -# Find the maximum coins you can collect by +# Find the maximum coins you can collect by # bursting the balloons wisely. # -# Note: +# Note: # (1) You may imagine nums[-1] = nums[n] = 1. # They are not real therefore you can not burst them. # (2) 0 <= n <= 500, 0 <= nums[i] <= 100 @@ -38,7 +38,7 @@ def maxCoins(self, nums): coins = [1] + [i for i in nums if i > 0] + [1] n = len(coins) max_coins = [[0 for _ in xrange(n)] for _ in xrange(n)] - + for k in xrange(2, n): for left in xrange(n - k): right = left + k @@ -48,4 +48,4 @@ def maxCoins(self, nums): max_coins[left][i] + max_coins[i][right]) return max_coins[0][-1] - + diff --git a/Python/can-i-win.py b/Python/can-i-win.py index 1877b04e9..8e482cf30 100644 --- a/Python/can-i-win.py +++ b/Python/can-i-win.py @@ -53,7 +53,7 @@ def canIWinHelper(maxChoosableInteger, desiredTotal, visited, lookup): mask <<= 1 lookup[visited] = False return False - + if (1 + maxChoosableInteger) * (maxChoosableInteger / 2) < desiredTotal: return False diff --git a/Python/candy-crush.py b/Python/candy-crush.py index 782f4af05..591398830 100644 --- a/Python/candy-crush.py +++ b/Python/candy-crush.py @@ -25,7 +25,7 @@ # # Example 1: # Input: -# board = +# board = # [[110,5,112,113,114],[210,211,5,213,214],[310,311,3,313,314],[410,411,412,5,414],[5,1,512,3,3],[610,4,1,613,614],[710,1,2,713,714],[810,1,2,1,1],[1,1,2,2,2],[4,1,4,4,1014]] # Output: # [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0],[110,0,0,0,114],[210,0,0,0,214],[310,0,0,113,314],[410,0,0,213,414],[610,211,112,313,614],[710,311,412,613,714],[810,411,512,713,1014]] @@ -43,7 +43,7 @@ def candyCrush(self, board): """ R, C = len(board), len(board[0]) changed = True - + while changed: changed = False diff --git a/Python/candy.py b/Python/candy.py index daf2f8153..eded6772a 100644 --- a/Python/candy.py +++ b/Python/candy.py @@ -2,9 +2,9 @@ # Space: O(n) # # There are N children standing in a line. Each child is assigned a rating value. -# +# # You are giving candies to these children subjected to the following requirements: -# +# # Each child must have at least one candy. # Children with a higher rating get more candies than their neighbors. # What is the minimum candies you must give? @@ -20,13 +20,13 @@ def candy(self, ratings): for i in xrange(1, len(ratings)): if ratings[i] > ratings[i - 1]: candies[i] = candies[i - 1] + 1 - + for i in reversed(xrange(1, len(ratings))): if ratings[i - 1] > ratings[i] and candies[i - 1] <= candies[i]: candies[i - 1] = candies[i] + 1 - + return reduce(operator.add, candies) - + if __name__ == "__main__": result = Solution().candy([1, 2, 3, 2, 3, 5, 2, 5]) print result diff --git a/Python/cherry-pickup.py b/Python/cherry-pickup.py index 4764f990d..f5815a5ab 100644 --- a/Python/cherry-pickup.py +++ b/Python/cherry-pickup.py @@ -20,7 +20,7 @@ # [1, 0, -1], # [1, 1, 1]] # Output: 5 -# Explanation: +# Explanation: # The player started at (0, 0) and went down, down, right right to reach (2, 2). # 4 cherries were picked up during this single trip, and the matrix becomes [[0,1,-1],[0,0,-1],[0,0,0]]. # Then, the player went left, up, up, left to return home, picking up one more cherry. @@ -38,7 +38,7 @@ def cherryPickup(self, grid): :rtype: int """ # dp holds the max # of cherries two k-length paths can pickup. - # The two k-length paths arrive at (i, k - i) and (j, k - j), + # The two k-length paths arrive at (i, k - i) and (j, k - j), # respectively. n = len(grid) dp = [[-1 for _ in xrange(n)] for _ in xrange(n)] diff --git a/Python/circular-array-loop.py b/Python/circular-array-loop.py index f1b9e10e4..32c7f87c7 100644 --- a/Python/circular-array-loop.py +++ b/Python/circular-array-loop.py @@ -26,18 +26,18 @@ def circularArrayLoop(self, nums): """ def next_index(nums, i): return (i + nums[i]) % len(nums) - + for i in xrange(len(nums)): if nums[i] == 0: continue - + slow, fast = i, i while nums[next_index(nums, slow)] * nums[i] > 0 and \ nums[next_index(nums, fast)] * nums[i] > 0 and \ nums[next_index(nums, next_index(nums, fast))] * nums[i] > 0: slow = next_index(nums, slow) fast = next_index(nums, next_index(nums, fast)) - if slow == fast: + if slow == fast: if slow == next_index(nums, slow): break return True diff --git a/Python/climbing-stairs.py b/Python/climbing-stairs.py index 6511a1dee..a6ccff427 100644 --- a/Python/climbing-stairs.py +++ b/Python/climbing-stairs.py @@ -2,8 +2,8 @@ # Space: O(1) # # You are climbing a stair case. It takes n steps to reach to the top. -# -# Each time you can either climb 1 or 2 steps. +# +# Each time you can either climb 1 or 2 steps. # In how many distinct ways can you climb to the top? @@ -15,7 +15,7 @@ class Solution: def climbStairs(self, n): prev, current = 0, 1 for i in xrange(n): - prev, current = current, prev + current, + prev, current = current, prev + current, return current # Time: O(2^n) diff --git a/Python/clone-graph.py b/Python/clone-graph.py index 87ecec311..a4a0b0338 100644 --- a/Python/clone-graph.py +++ b/Python/clone-graph.py @@ -2,21 +2,21 @@ # Space: O(n) # # Clone an undirected graph. Each node in the graph contains a label and a list of its neighbors. -# -# +# +# # OJ's undirected graph serialization: # Nodes are labeled uniquely. -# +# # We use # as a separator for each node, and , as a separator for node label and each neighbor of the node. # As an example, consider the serialized graph {0,1,2#1,2#2,2}. -# +# # The graph has a total of three nodes, and therefore contains three parts as separated by #. -# +# # First node is labeled as 0. Connect node 0 to both nodes 1 and 2. # Second node is labeled as 1. Connect node 1 to node 2. # Third node is labeled as 2. Connect node 2 to node 2 (itself), thus forming a self-cycle. # Visually, the graph looks like the following: -# +# # 1 # / \ # / \ @@ -38,7 +38,7 @@ def cloneGraph(self, node): return None cloned_node = UndirectedGraphNode(node.label) cloned, queue = {node:cloned_node}, [node] - + while queue: current = queue.pop() for neighbor in current.neighbors: diff --git a/Python/closest-binary-search-tree-value-ii.py b/Python/closest-binary-search-tree-value-ii.py index 4672e0cc6..3abd38a54 100644 --- a/Python/closest-binary-search-tree-value-ii.py +++ b/Python/closest-binary-search-tree-value-ii.py @@ -27,11 +27,11 @@ def nextNode(stack, child1, child2): child = stack.pop() while stack and child is child2(stack): child = stack.pop() - + # The forward or backward iterator. backward = lambda stack: stack[-1].left forward = lambda stack: stack[-1].right - + # Build the stack to the closest node. stack = [] while root: @@ -39,11 +39,11 @@ def nextNode(stack, child1, child2): root = root.left if target < root.val else root.right dist = lambda node: abs(node.val - target) forward_stack = stack[:stack.index(min(stack, key=dist))+1] - + # Get the stack to the next smaller node. backward_stack = list(forward_stack) nextNode(backward_stack, backward, forward) - + # Get the closest k values by advancing the iterators of the stacks. result = [] for _ in xrange(k): @@ -74,7 +74,7 @@ def __init__(self, stack, child1, child2): self.cur = self.stack.pop() self.child1 = child1 self.child2 = child2 - + # @return an integer, the next node def next(self): node = None @@ -103,7 +103,7 @@ def next(self): root = root.left if target < root.val else root.right dist = lambda node: abs(node.val - target) if node else float("inf") stack = stack[:stack.index(min(stack, key=dist))+1] - + # The forward or backward iterator. backward = lambda node: node.left forward = lambda node: node.right diff --git a/Python/coin-change-2.py b/Python/coin-change-2.py index d65707685..c85c9d196 100644 --- a/Python/coin-change-2.py +++ b/Python/coin-change-2.py @@ -27,9 +27,9 @@ # Explanation: the amount of 3 cannot be made up just with coins of 2. # Example 3: # -# Input: amount = 10, coins = [10] +# Input: amount = 10, coins = [10] # Output: 1 - + class Solution(object): def change(self, amount, coins): """ diff --git a/Python/coin-change.py b/Python/coin-change.py index f762c382a..093d4429b 100644 --- a/Python/coin-change.py +++ b/Python/coin-change.py @@ -35,4 +35,4 @@ def coinChange(self, coins, amount): if i + coin <= amount: amounts[i + coin] = min(amounts[i + coin], amounts[i] + 1) return amounts[amount] if amounts[amount] != INF else -1 - + diff --git a/Python/combination-sum-ii.py b/Python/combination-sum-ii.py index d11d74295..31975069d 100644 --- a/Python/combination-sum-ii.py +++ b/Python/combination-sum-ii.py @@ -1,20 +1,20 @@ # Time: O(k * C(n, k)) # Space: O(k) -# -# Given a collection of candidate numbers (C) and a target number (T), +# +# Given a collection of candidate numbers (C) and a target number (T), # find all unique combinations in C where the candidate numbers sums to T. -# +# # Each number in C may only be used once in the combination. -# +# # Note: # All numbers (including target) will be positive integers. # Elements in a combination (a1, a2, ... , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak). # The solution set must not contain duplicate combinations. -# For example, given candidate set 10,1,2,7,6,1,5 and target 8, -# A solution set is: -# [1, 7] -# [1, 2, 5] -# [2, 6] +# For example, given candidate set 10,1,2,7,6,1,5 and target 8, +# A solution set is: +# [1, 7] +# [1, 2, 5] +# [2, 6] # [1, 1, 6] # @@ -26,7 +26,7 @@ def combinationSum2(self, candidates, target): result = [] self.combinationSumRecu(sorted(candidates), result, 0, [], target) return result - + def combinationSumRecu(self, candidates, result, start, intermediate, target): if target == 0: result.append(list(intermediate)) @@ -41,5 +41,5 @@ def combinationSumRecu(self, candidates, result, start, intermediate, target): if __name__ == "__main__": candidates, target = [10, 1, 2, 7, 6, 1, 5], 8 - result = Solution().combinationSum2(candidates, target) - print result + result = Solution().combinationSum2(candidates, target) + print result diff --git a/Python/combination-sum-iii.py b/Python/combination-sum-iii.py index ddc4ba260..f5accf25a 100644 --- a/Python/combination-sum-iii.py +++ b/Python/combination-sum-iii.py @@ -32,7 +32,7 @@ def combinationSum3(self, k, n): result = [] self.combinationSumRecu(result, [], 1, k, n) return result - + def combinationSumRecu(self, result, intermediate, start, k, target): if k == 0 and target == 0: result.append(list(intermediate)) diff --git a/Python/combination-sum.py b/Python/combination-sum.py index ce0d14f11..0120a7cdb 100644 --- a/Python/combination-sum.py +++ b/Python/combination-sum.py @@ -1,19 +1,19 @@ # Time: O(k * n^k) # Space: O(k) # -# Given a set of candidate numbers (C) and a target number (T), +# Given a set of candidate numbers (C) and a target number (T), # find all unique combinations in C where the candidate numbers sums to T. -# +# # The same repeated number may be chosen from C unlimited number of times. -# +# # Note: # All numbers (including target) will be positive integers. # Elements in a combination (a1, a2, ... , ak) must be in non-descending order. (ie, a1 <= a2 <= ... <= ak). # The solution set must not contain duplicate combinations. -# For example, given candidate set 2,3,6,7 and target 7, -# A solution set is: -# [7] -# [2, 2, 3] +# For example, given candidate set 2,3,6,7 and target 7, +# A solution set is: +# [7] +# [2, 2, 3] # class Solution: @@ -24,7 +24,7 @@ def combinationSum(self, candidates, target): result = [] self.combinationSumRecu(sorted(candidates), result, 0, [], target) return result - + def combinationSumRecu(self, candidates, result, start, intermediate, target): if target == 0: result.append(list(intermediate)) @@ -36,5 +36,5 @@ def combinationSumRecu(self, candidates, result, start, intermediate, target): if __name__ == "__main__": candidates, target = [2, 3, 6, 7], 7 - result = Solution().combinationSum(candidates, target) - print result + result = Solution().combinationSum(candidates, target) + print result diff --git a/Python/combinations.py b/Python/combinations.py index 7a4ba2a5f..e57640ab1 100644 --- a/Python/combinations.py +++ b/Python/combinations.py @@ -2,10 +2,10 @@ # Space: O(k) # Given two integers n and k, return all possible combinations of k numbers out of 1 ... n. -# +# # For example, # If n = 4 and k = 2, a solution is: -# +# # [ # [2,4], # [3,4], @@ -36,7 +36,7 @@ def combine(self, n, k): combination.append(i) i += 1 return result - + class Solution2(object): def combine(self, n, k): diff --git a/Python/compare-version-numbers.py b/Python/compare-version-numbers.py index 4b67b70e4..342dff213 100644 --- a/Python/compare-version-numbers.py +++ b/Python/compare-version-numbers.py @@ -4,17 +4,17 @@ # Compare two version numbers version1 and version1. # If version1 > version2 return 1, if version1 < version2 # return -1, otherwise return 0. -# +# # You may assume that the version strings are non-empty and # contain only digits and the . character. -# The . character does not represent a decimal point and +# The . character does not represent a decimal point and # is used to separate number sequences. # For instance, 2.5 is not "two and a half" or "half way to # version three", it is the fifth second-level revision of # the second first-level revision. -# +# # Here is an example of version numbers ordering: -# +# # 0.1 < 1.1 < 1.2 < 13.37 # import itertools @@ -56,12 +56,12 @@ def compareVersion(self, version1, version2): :rtype: int """ v1, v2 = version1.split("."), version2.split(".") - + if len(v1) > len(v2): v2 += ['0' for _ in xrange(len(v1) - len(v2))] elif len(v1) < len(v2): v1 += ['0' for _ in xrange(len(v2) - len(v1))] - + i = 0 while i < len(v1): if int(v1[i]) > int(v2[i]): @@ -70,7 +70,7 @@ def compareVersion(self, version1, version2): return -1 else: i += 1 - + return 0 def compareVersion2(self, version1, version2): diff --git a/Python/concatenated-words.py b/Python/concatenated-words.py index 8d89f5e36..6fceb631c 100644 --- a/Python/concatenated-words.py +++ b/Python/concatenated-words.py @@ -12,8 +12,8 @@ # # Output: ["catsdogcats","dogcatsdog","ratcatdogcat"] # -# Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; -# "dogcatsdog" can be concatenated by "dog", "cats" and "dog"; +# Explanation: "catsdogcats" can be concatenated by "cats", "dog" and "cats"; +# "dogcatsdog" can be concatenated by "dog", "cats" and "dog"; # "ratcatdogcat" can be concatenated by "rat", "cat", "dog" and "cat". # Note: # The number of elements of the given array will not exceed 10,000 @@ -29,15 +29,15 @@ def findAllConcatenatedWordsInADict(self, words): """ lookup = set(words) result = [] - for word in words: + for word in words: dp = [False] * (len(word)+1) dp[0] = True for i in xrange(len(word)): if not dp[i]: continue - + for j in xrange(i+1, len(word)+1): - if j - i < len(word) and word[i:j] in lookup: + if j - i < len(word) and word[i:j] in lookup: dp[j] = True if dp[len(word)]: diff --git a/Python/construct-binary-tree-from-inorder-and-postorder-traversal.py b/Python/construct-binary-tree-from-inorder-and-postorder-traversal.py index 991009a93..294594fa5 100644 --- a/Python/construct-binary-tree-from-inorder-and-postorder-traversal.py +++ b/Python/construct-binary-tree-from-inorder-and-postorder-traversal.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given inorder and postorder traversal of a tree, construct the binary tree. -# +# # Note: # You may assume that duplicates do not exist in the tree. # @@ -23,7 +23,7 @@ def buildTree(self, inorder, postorder): for i, num in enumerate(inorder): lookup[num] = i return self.buildTreeRecu(lookup, postorder, inorder, len(postorder), 0, len(inorder)) - + def buildTreeRecu(self, lookup, postorder, inorder, post_end, in_start, in_end): if in_start == in_end: return None diff --git a/Python/construct-binary-tree-from-preorder-and-inorder-traversal.py b/Python/construct-binary-tree-from-preorder-and-inorder-traversal.py index e3196454d..a316ecdd2 100644 --- a/Python/construct-binary-tree-from-preorder-and-inorder-traversal.py +++ b/Python/construct-binary-tree-from-preorder-and-inorder-traversal.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given preorder and inorder traversal of a tree, construct the binary tree. -# +# # Note: # You may assume that duplicates do not exist in the tree. # @@ -23,7 +23,7 @@ def buildTree(self, preorder, inorder): for i, num in enumerate(inorder): lookup[num] = i return self.buildTreeRecu(lookup, preorder, inorder, 0, 0, len(inorder)) - + def buildTreeRecu(self, lookup, preorder, inorder, pre_start, in_start, in_end): if in_start == in_end: return None diff --git a/Python/construct-string-from-binary-tree.py b/Python/construct-string-from-binary-tree.py index 8bba46ba5..b2dac24ee 100644 --- a/Python/construct-string-from-binary-tree.py +++ b/Python/construct-string-from-binary-tree.py @@ -5,7 +5,7 @@ # the preorder traversing way. # # The null node needs to be represented by empty parenthesis pair "()". -# And you need to omit all the empty parenthesis pairs that don't affect +# And you need to omit all the empty parenthesis pairs that don't affect # the one-to-one mapping relationship between the string and the original binary tree. # # Example 1: @@ -13,25 +13,25 @@ # 1 # / \ # 2 3 -# / -# 4 +# / +# 4 # # Output: "1(2(4))(3)" # -# Explanation: Originallay it needs to be "1(2(4)())(3()())", -# but you need to omit all the unnecessary empty parenthesis pairs. +# Explanation: Originallay it needs to be "1(2(4)())(3()())", +# but you need to omit all the unnecessary empty parenthesis pairs. # And it will be "1(2(4))(3)". # Example 2: # Input: Binary tree: [1,2,3,null,4] # 1 # / \ # 2 3 -# \ -# 4 +# \ +# 4 # # Output: "1(2()(4))(3)" # -# Explanation: Almost the same as the first example, +# Explanation: Almost the same as the first example, # except we can't omit the first parenthesis pair to break the one-to-one mapping relationship # between the input and the output. diff --git a/Python/contain-virus.py b/Python/contain-virus.py index abf74ca92..3fc0e7569 100644 --- a/Python/contain-virus.py +++ b/Python/contain-virus.py @@ -11,7 +11,7 @@ # # Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall. # Resources are limited. -# Each day, you can install walls around only one region -- +# Each day, you can install walls around only one region -- # the affected area (continuous block of infected cells) that # threatens the most uninfected cells the following night. # There will never be a tie. @@ -20,7 +20,7 @@ # If not, and the world becomes fully infected, return the number of walls used. # # Example 1: -# Input: grid = +# Input: grid = # [[0,1,0,0,0,0,0,1], # [0,1,0,0,0,0,0,1], # [0,0,0,0,0,0,0,1], @@ -38,7 +38,7 @@ # On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained. # # Example 2: -# Input: grid = +# Input: grid = # [[1,1,1], # [1,0,1], # [1,1,1]] @@ -47,7 +47,7 @@ # Notice that walls are only built on the shared boundary of two different cells. # # Example 3: -# Input: grid = +# Input: grid = # [[1,1,1,0,0,0,0,0,0], # [1,0,1,0,1,1,1,1,1], # [1,1,1,0,0,0,0,0,0]] diff --git a/Python/container-with-most-water.py b/Python/container-with-most-water.py index aa81a7018..5522e0080 100644 --- a/Python/container-with-most-water.py +++ b/Python/container-with-most-water.py @@ -1,13 +1,13 @@ # Time: O(n) # Space: O(1) # -# Given n non-negative integers a1, a2, ..., an, -# where each represents a point at coordinate (i, ai). -# n vertical lines are drawn such that the two endpoints of -# line i is at (i, ai) and (i, 0). Find two lines, -# which together with x-axis forms a container, +# Given n non-negative integers a1, a2, ..., an, +# where each represents a point at coordinate (i, ai). +# n vertical lines are drawn such that the two endpoints of +# line i is at (i, ai) and (i, 0). Find two lines, +# which together with x-axis forms a container, # such that the container contains the most water. -# +# # Note: You may not slant the container. # @@ -22,7 +22,7 @@ def maxArea(self, height): else: j -= 1 return max_area - + if __name__ == "__main__": height = [1, 2, 3, 4, 3, 2, 1, 5] result = Solution().maxArea(height) diff --git a/Python/contains-duplicate-ii.py b/Python/contains-duplicate-ii.py index 451a6bdde..ca5206a55 100644 --- a/Python/contains-duplicate-ii.py +++ b/Python/contains-duplicate-ii.py @@ -1,8 +1,8 @@ # Time: O(n) # Space: O(n) # -# Given an array of integers and an integer k, return true if -# and only if there are two distinct indices i and j in the array +# Given an array of integers and an integer k, return true if +# and only if there are two distinct indices i and j in the array # such that nums[i] = nums[j] and the difference between i and j is at most k. # diff --git a/Python/contains-duplicate.py b/Python/contains-duplicate.py index 16c26a3c3..a6ba14224 100644 --- a/Python/contains-duplicate.py +++ b/Python/contains-duplicate.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given an array of integers, find if the array contains any duplicates. -# Your function should return true if any value appears at least twice in the array, +# Your function should return true if any value appears at least twice in the array, # and it should return false if every element is distinct. # diff --git a/Python/contiguous-array.py b/Python/contiguous-array.py index f16e9147d..104194c0b 100644 --- a/Python/contiguous-array.py +++ b/Python/contiguous-array.py @@ -27,5 +27,5 @@ def findMaxLength(self, nums): result = max(result, i - lookup[count]) else: lookup[count] = i - + return result diff --git a/Python/continuous-subarray-sum.py b/Python/continuous-subarray-sum.py index b9991538f..7e46a53bc 100644 --- a/Python/continuous-subarray-sum.py +++ b/Python/continuous-subarray-sum.py @@ -36,5 +36,5 @@ def checkSubarraySum(self, nums, k): return True else: lookup[count] = i - + return False diff --git a/Python/convert-bst-to-greater-tree.py b/Python/convert-bst-to-greater-tree.py index c0ff15309..5262d45dd 100644 --- a/Python/convert-bst-to-greater-tree.py +++ b/Python/convert-bst-to-greater-tree.py @@ -17,7 +17,7 @@ # 18 # / \ # 20 13 - + # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): diff --git a/Python/convert-sorted-array-to-binary-search-tree.py b/Python/convert-sorted-array-to-binary-search-tree.py index 4f5858c07..ac3789fb5 100644 --- a/Python/convert-sorted-array-to-binary-search-tree.py +++ b/Python/convert-sorted-array-to-binary-search-tree.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(logn) # -# Given an array where elements are sorted in ascending order, +# Given an array where elements are sorted in ascending order, # convert it to a height balanced BST. # # Definition for a binary tree node @@ -28,7 +28,7 @@ def sortedArrayToBSTRecu(self, nums, start, end): node.left = self.sortedArrayToBSTRecu(nums, start, mid) node.right = self.sortedArrayToBSTRecu(nums, mid + 1, end) return node - + def perfect_tree_pivot(self, n): """ Find the point to partition n keys for a perfect binary search tree @@ -44,8 +44,8 @@ def perfect_tree_pivot(self, n): else: return n - x // 2 # case 2 == n - (x//2 - 1) - 1 : the left subtree of the root # has more nodes and the right subtree is perfect. - - + + if __name__ == "__main__": num = [1, 2, 3] result = Solution().sortedArrayToBST(num) diff --git a/Python/convert-sorted-list-to-binary-search-tree.py b/Python/convert-sorted-list-to-binary-search-tree.py index f0624af6c..a1e2a5166 100644 --- a/Python/convert-sorted-list-to-binary-search-tree.py +++ b/Python/convert-sorted-list-to-binary-search-tree.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(logn) # -# Given a singly linked list where elements are sorted in ascending order, +# Given a singly linked list where elements are sorted in ascending order, # convert it to a height balanced BST. # # Definition for a binary tree node @@ -27,14 +27,14 @@ def sortedListToBST(self, head): current, length = current.next, length + 1 self.head = head return self.sortedListToBSTRecu(0, length) - + def sortedListToBSTRecu(self, start, end): if start == end: return None mid = start + (end - start) / 2 left = self.sortedListToBSTRecu(start, mid) current = TreeNode(self.head.val) - current.left = left + current.left = left self.head = self.head.next current.right = self.sortedListToBSTRecu(mid + 1, end) return current diff --git a/Python/convex-polygon.py b/Python/convex-polygon.py index 27286efd6..a552142c3 100644 --- a/Python/convex-polygon.py +++ b/Python/convex-polygon.py @@ -19,4 +19,4 @@ def det(A): return False prev = curr return True - + diff --git a/Python/copy-list-with-random-pointer.py b/Python/copy-list-with-random-pointer.py index 9f4e8675f..483aa2a3e 100644 --- a/Python/copy-list-with-random-pointer.py +++ b/Python/copy-list-with-random-pointer.py @@ -3,7 +3,7 @@ # # A linked list is given such that each node contains an additional random pointer # which could point to any node in the list or null. -# +# # Return a deep copy of the list. # @@ -25,14 +25,14 @@ def copyRandomList(self, head): copied.next = current.next current.next = copied current = copied.next - + # update random node in copied list current = head while current: if current.random: current.next.random = current.random.next current = current.next.next - + # split copied list from combined one dummy = RandomListNode(0) copied_current, current = dummy, head @@ -50,19 +50,19 @@ class Solution2: def copyRandomList(self, head): dummy = RandomListNode(0) current, prev, copies = head, dummy, {} - + while current: copied = RandomListNode(current.label) copies[current] = copied prev.next = copied prev, current = prev.next, current.next - + current = head while current: if current.random: copies[current].random = copies[current.random] current = current.next - + return dummy.next diff --git a/Python/count-and-say.py b/Python/count-and-say.py index f1b27c374..2bbfc3a4e 100644 --- a/Python/count-and-say.py +++ b/Python/count-and-say.py @@ -3,12 +3,12 @@ # # The count-and-say sequence is the sequence of integers beginning as follows: # 1, 11, 21, 1211, 111221, ... -# +# # 1 is read off as "one 1" or 11. # 11 is read off as "two 1s" or 21. # 21 is read off as "one 2, then one 1" or 1211. # Given an integer n, generate the nth sequence. -# +# # Note: The sequence of integers will be represented as a string. # @@ -19,11 +19,11 @@ def countAndSay(self, n): for i in xrange(n - 1): seq = self.getNext(seq) return seq - + def getNext(self, seq): i, next_seq = 0, "" while i < len(seq): - cnt = 1 + cnt = 1 while i < len(seq) - 1 and seq[i] == seq[i + 1]: cnt += 1 i += 1 @@ -35,4 +35,4 @@ def getNext(self, seq): for i in xrange(1, 4): print Solution().countAndSay(i) - + diff --git a/Python/count-binary-substrings.py b/Python/count-binary-substrings.py index f282b349a..e28ba957a 100644 --- a/Python/count-binary-substrings.py +++ b/Python/count-binary-substrings.py @@ -20,7 +20,7 @@ # Input: "10101" # Output: 4 # Explanation: There are 4 substrings: "10", "01", "10", "01" that have equal number of consecutive 1's and 0's. -# +# # Note: # s.length will be between 1 and 50,000. # s will only consist of "0" or "1" characters. diff --git a/Python/count-complete-tree-nodes.py b/Python/count-complete-tree-nodes.py index 24695387b..dcd6e490a 100644 --- a/Python/count-complete-tree-nodes.py +++ b/Python/count-complete-tree-nodes.py @@ -22,12 +22,12 @@ class Solution: def countNodes(self, root): if root is None: return 0 - + node, level = root, 0 while node.left is not None: node = node.left level += 1 - + # Binary search. left, right = 2 ** level, 2 ** (level + 1) while left < right: @@ -36,16 +36,16 @@ def countNodes(self, root): right = mid else: left = mid + 1 - + return left - 1 - + # Check if the nth node exist. def exist(self, root, n): k = 1 while k <= n: k <<= 1 k >>= 2 - + node = root while k > 0: if (n & k) == 0: diff --git a/Python/count-different-palindromic-subsequences.py b/Python/count-different-palindromic-subsequences.py index e0b4e7452..f1b5ecf1e 100644 --- a/Python/count-different-palindromic-subsequences.py +++ b/Python/count-different-palindromic-subsequences.py @@ -11,19 +11,19 @@ # Two sequences A_1, A_2, ... and B_1, B_2, ... are different if there is some i for which A_i != B_i. # # Example 1: -# Input: +# Input: # S = 'bccb' # Output: 6 -# Explanation: +# Explanation: # The 6 different non-empty palindromic subsequences are 'b', 'c', 'bb', 'cc', 'bcb', 'bccb'. # Note that 'bcb' is counted only once, even though it occurs twice. # # Example 2: -# Input: +# Input: # S = 'abcdabcdabcdabcdabcdabcdabcdabcddcbadcbadcbadcbadcbadcbadcbadcba' # Output: 104860361 -# -# Explanation: +# +# Explanation: # There are 3104860382 different non-empty palindromic subsequences, which is 104860361 modulo 10^9 + 7. # Note: # - The length of S will be in the range [1, 1000]. @@ -50,20 +50,20 @@ def dp(i, j, prv, nxt, lookup): result %= P lookup[i][j] = result return result - + prv = [None] * len(S) nxt = [None] * len(S) - + last = [None] * 4 for i in xrange(len(S)): last[ord(S[i])-ord('a')] = i prv[i] = tuple(last) - + last = [None] * 4 for i in reversed(xrange(len(S))): last[ord(S[i])-ord('a')] = i nxt[i] = tuple(last) - + P = 10**9 + 7 - lookup = [[None] * len(S) for _ in xrange(len(S))] + lookup = [[None] * len(S) for _ in xrange(len(S))] return dp(0, len(S)-1, prv, nxt, lookup) - 1 diff --git a/Python/count-of-range-sum.py b/Python/count-of-range-sum.py index eb87af804..01242acc0 100644 --- a/Python/count-of-range-sum.py +++ b/Python/count-of-range-sum.py @@ -39,7 +39,7 @@ def countAndMergeSort(sums, start, end, lower, upper): while j < end and sums[j] - sums[i] <= upper: j += 1 count += j - k - + # Merge the two sorted arrays into tmp. while r < end and sums[r] < sums[i]: tmp.append(sums[r]) @@ -80,7 +80,7 @@ def countAndMergeSort(sums, start, end, lower, upper): while j <= end and sums[j] - sums[i] <= upper: j += 1 count += j - k - + # Merge the two sorted arrays into tmp. while r <= end and sums[r] < sums[i]: tmp.append(sums[r]) diff --git a/Python/count-of-smaller-numbers-after-self.py b/Python/count-of-smaller-numbers-after-self.py index 7011fa583..42f49c324 100644 --- a/Python/count-of-smaller-numbers-after-self.py +++ b/Python/count-of-smaller-numbers-after-self.py @@ -1,9 +1,9 @@ # Time: O(nlogn) # Space: O(n) -# You are given an integer array nums and you have to -# return a new counts array. The counts array has the -# property where counts[i] is the number of smaller +# You are given an integer array nums and you have to +# return a new counts array. The counts array has the +# property where counts[i] is the number of smaller # elements to the right of nums[i]. # # Example: @@ -77,7 +77,7 @@ def add(self, i, val): while i < len(self.__bit): self.__bit[i] += val i += (i & -i) - + def query(self, i): ret = 0 while i > 0: @@ -114,17 +114,17 @@ def countSmaller(self, nums): res[i] = bst.query(nums[i]) return res - + class BST(object): class BSTreeNode(object): def __init__(self, val): self.val = val self.count = 0 self.left = self.right = None - + def __init__(self): self.root = None - + # Insert node into BST. def insertNode(self, val): node = self.BSTreeNode(val) @@ -147,7 +147,7 @@ def insertNode(self, val): else: curr.right = node break - + # Query the smaller count of the value. def query(self, val): count = 0 @@ -160,5 +160,5 @@ def query(self, val): count += 1 + curr.count # Count the number of the smaller nodes. curr = curr.right else: # Equal. - return count + curr.count + return count + curr.count return 0 diff --git a/Python/count-primes.py b/Python/count-primes.py index 46ad88ddf..569abf3d0 100644 --- a/Python/count-primes.py +++ b/Python/count-primes.py @@ -14,7 +14,7 @@ class Solution: def countPrimes(self, n): if n <= 2: return 0 - + is_prime = [True] * n num = n / 2 for i in xrange(3, n, 2): diff --git a/Python/count-the-repetitions.py b/Python/count-the-repetitions.py index eeaee7b70..37d8b47cf 100644 --- a/Python/count-the-repetitions.py +++ b/Python/count-the-repetitions.py @@ -4,7 +4,7 @@ # Define S = [s,n] as the string S which consists of n connected strings s. # For example, ["abc", 3] ="abcabcabc". # -# On the other hand, we define that string s1 can be obtained from string s2 +# On the other hand, we define that string s1 can be obtained from string s2 # if we can remove some characters from s2 such that it becomes s1. # For example, “abc” can be obtained from “abdbec” based on our definition, but it can not be obtained from “acbbe”. # @@ -38,7 +38,7 @@ def getMaxRepetitions(self, s1, n1, s2, n2): if s1[i] == s2[j]: j = (j + 1) % len(s2) count += (j == 0) - + if j in lookup: # cyclic i = lookup[j] prefix_count = repeat_count[i] @@ -47,5 +47,5 @@ def getMaxRepetitions(self, s1, n1, s2, n2): return (prefix_count + pattern_count + suffix_count) / n2 lookup[j] = k repeat_count[k] = count - + return repeat_count[n1] / n2 # not cyclic iff n1 <= s2 diff --git a/Python/count-univalue-subtrees.py b/Python/count-univalue-subtrees.py index a393f4304..c70a5ba5d 100644 --- a/Python/count-univalue-subtrees.py +++ b/Python/count-univalue-subtrees.py @@ -27,6 +27,6 @@ def isUnivalSubtrees(self, root, count): return [True, count] return [False, count] - + def isSame(self, root, child, is_uni): return not child or (is_uni and root.val == child.val) diff --git a/Python/counting-bits.py b/Python/counting-bits.py index 6ef845380..8da8a2185 100644 --- a/Python/counting-bits.py +++ b/Python/counting-bits.py @@ -2,7 +2,7 @@ # Space: O(n) # Given a non negative integer number num. For every numbers i -# in the range 0 <= i <= num calculate the number +# in the range 0 <= i <= num calculate the number # of 1's in their binary representation and return them as an array. # # Example: @@ -10,8 +10,8 @@ # # Follow up: # -# It is very easy to come up with a solution with run -# time O(n*sizeof(integer)). But can you do it in +# It is very easy to come up with a solution with run +# time O(n*sizeof(integer)). But can you do it in # linear time O(n) /possibly in a single pass? # Space complexity should be O(n). # Can you do it like a boss? Do it without using diff --git a/Python/couples-holding-hands.py b/Python/couples-holding-hands.py index 2348c41c8..0801bcd81 100644 --- a/Python/couples-holding-hands.py +++ b/Python/couples-holding-hands.py @@ -15,7 +15,7 @@ def minSwapsCouples(self, row): for couch1, couch2 in couples: adj[couch1].append(couch2) adj[couch2].append(couch1) - + result = 0 for couch in xrange(N): if not adj[couch]: continue diff --git a/Python/course-schedule.py b/Python/course-schedule.py index ecb7fbf72..ddd5fc3de 100644 --- a/Python/course-schedule.py +++ b/Python/course-schedule.py @@ -2,27 +2,27 @@ # Space: O(|E|) # # There are a total of n courses you have to take, labeled from 0 to n - 1. -# -# Some courses may have prerequisites, for example to take course 0 +# +# Some courses may have prerequisites, for example to take course 0 # you have to first take course 1, which is expressed as a pair: [0,1] -# +# # Given the total number of courses and a list of prerequisite pairs, # is it possible for you to finish all courses? -# +# # For example: -# +# # 2, [[1,0]] -# There are a total of 2 courses to take. To take course 1 +# There are a total of 2 courses to take. To take course 1 # you should have finished course 0. So it is possible. -# +# # 2, [[1,0],[0,1]] -# There are a total of 2 courses to take. To take course 1 you should have +# There are a total of 2 courses to take. To take course 1 you should have # finished course 0, and to take course 0 you should also have finished course 1. So it is impossible. -# +# # click to show more hints. -# +# # Hints: -# This problem is equivalent to finding if a cycle exists in a directed graph. +# This problem is equivalent to finding if a cycle exists in a directed graph. # If a cycle exists, no topological ordering exists and therefore it will be impossible to take all courses. # There are several ways to represent a graph. For example, the input prerequisites is a graph represented by # a list of edges. Is this graph representation appropriate? @@ -41,7 +41,7 @@ def canFinish(self, numCourses, prerequisites): :rtype: bool """ zero_in_degree_queue, in_degree, out_degree = collections.deque(), {}, {} - + for i, j in prerequisites: if i not in in_degree: in_degree[i] = set() @@ -49,25 +49,25 @@ def canFinish(self, numCourses, prerequisites): out_degree[j] = set() in_degree[i].add(j) out_degree[j].add(i) - + for i in xrange(numCourses): if i not in in_degree: zero_in_degree_queue.append(i) - + while zero_in_degree_queue: prerequisite = zero_in_degree_queue.popleft() - + if prerequisite in out_degree: for course in out_degree[prerequisite]: in_degree[course].discard(prerequisite) if not in_degree[course]: zero_in_degree_queue.append(course) - + del out_degree[prerequisite] - + if out_degree: return False - + return True diff --git a/Python/cracking-the-safe.py b/Python/cracking-the-safe.py index 916fc26fc..e4874bfcd 100644 --- a/Python/cracking-the-safe.py +++ b/Python/cracking-the-safe.py @@ -16,12 +16,12 @@ # Input: n = 1, k = 2 # Output: "01" # Note: "10" will be accepted too. -# +# # Example 2: # Input: n = 2, k = 2 # Output: "00110" # Note: "01100", "10011", "11001" will be accepted too. -# +# # Note: # - n will be in the range [1, 4]. # - k will be in the range [1, 10]. @@ -88,7 +88,7 @@ def dfs(k, node, lookup, result): result.append(str(i)) dfs(k, neighbor[1:], lookup, result) break - + result = [str(k-1)]*(n-1) lookup = set() dfs(k, "".join(result), lookup, result) diff --git a/Python/create-maximum-number.py b/Python/create-maximum-number.py index af1d98541..c946b12fa 100644 --- a/Python/create-maximum-number.py +++ b/Python/create-maximum-number.py @@ -57,7 +57,7 @@ def delete_digit(nums): res = res[:i] + res[i+1:] break return res - + def merge(a, b): return [max(a, b).pop(0) for _ in xrange(len(a)+len(b))] diff --git a/Python/daily-temperatures.py b/Python/daily-temperatures.py index 841982d81..49e55ab96 100644 --- a/Python/daily-temperatures.py +++ b/Python/daily-temperatures.py @@ -25,4 +25,4 @@ def dailyTemperatures(self, temperatures): idx = stk.pop() result[idx] = i-idx stk.append(i) - return result + return result diff --git a/Python/decode-string.py b/Python/decode-string.py index ab0256f17..c3030c3b5 100644 --- a/Python/decode-string.py +++ b/Python/decode-string.py @@ -3,8 +3,8 @@ # Given an encoded string, return it's decoded string. # -# The encoding rule is: k[encoded_string], -# where the encoded_string inside the square brackets is +# The encoding rule is: k[encoded_string], +# where the encoded_string inside the square brackets is # being repeated exactly k times. Note that k is guaranteed # to be a positive integer. # diff --git a/Python/decode-ways.py b/Python/decode-ways.py index 2f474480b..7d60cfa09 100644 --- a/Python/decode-ways.py +++ b/Python/decode-ways.py @@ -2,16 +2,16 @@ # Space: O(1) # # A message containing letters from A-Z is being encoded to numbers using the following mapping: -# +# # 'A' -> 1 # 'B' -> 2 # ... # 'Z' -> 26 # Given an encoded message containing digits, determine the total number of ways to decode it. -# +# # For example, # Given encoded message "12", it could be decoded as "AB" (1 2) or "L" (12). -# +# # The number of ways decoding "12" is 2. # diff --git a/Python/delete-and-earn.py b/Python/delete-and-earn.py index 4327d0d2b..4f4c82aba 100644 --- a/Python/delete-and-earn.py +++ b/Python/delete-and-earn.py @@ -12,14 +12,14 @@ # Example 1: # Input: nums = [3, 4, 2] # Output: 6 -# Explanation: +# Explanation: # Delete 4 to earn 4 points, consequently 3 is also deleted. # Then, delete 2 to earn 2 points. 6 total points are earned. # # Example 2: # Input: nums = [2, 2, 3, 3, 3, 4] # Output: 9 -# Explanation: +# Explanation: # Delete 3 to earn 3 points, deleting both 2's and the 4. # Then, delete 3 again to earn 3 points, and 3 again to earn 3 points. # 9 total points are earned. diff --git a/Python/design-in-memory-file-system.py b/Python/design-in-memory-file-system.py index ae5ee030c..0785bfff0 100644 --- a/Python/design-in-memory-file-system.py +++ b/Python/design-in-memory-file-system.py @@ -24,7 +24,7 @@ # readContentFromFile: Given a file path, return its content in string format. # # Example: -# Input: +# Input: # ["FileSystem","ls","mkdir","addContentToFile","ls","readContentFromFile"] # [[],["/"],["/a/b/c"],["/a/b/c/d","hello"],["/"],["/a/b/c/d"]] # Output: @@ -37,10 +37,10 @@ # users will not attempt to retrieve file content or list a directory or file that does not exist. # 3. You can assume that all directory names and file names only contain lower-case letters, # and same names won't exist in the same directory. - + class TrieNode(object): - + def __init__(self): self.is_file = False self.children = {} @@ -50,7 +50,7 @@ class FileSystem(object): def __init__(self): self.__root = TrieNode() - + def ls(self, path): """ @@ -58,11 +58,11 @@ def ls(self, path): :rtype: List[str] """ curr = self.__getNode(path) - + if curr.is_file: return [self.__split(path, '/')[-1]] - return sorted(curr.children.keys()) + return sorted(curr.children.keys()) def mkdir(self, path): @@ -72,7 +72,7 @@ def mkdir(self, path): """ curr = self.__putNode(path) curr.is_file = False - + def addContentToFile(self, filePath, content): """ diff --git a/Python/design-log-storage-system.py b/Python/design-log-storage-system.py index 976a44f50..ca9a90979 100644 --- a/Python/design-log-storage-system.py +++ b/Python/design-log-storage-system.py @@ -2,14 +2,14 @@ # retrieve: O(n + dlogd), n is the size of the total logs # , d is the size of the found logs # Space: O(n) - + class LogSystem(object): def __init__(self): self.__logs = [] self.__granularity = {'Year': 4, 'Month': 7, 'Day': 10, \ 'Hour': 13, 'Minute': 16, 'Second': 19} - + def put(self, id, timestamp): """ @@ -19,7 +19,7 @@ def put(self, id, timestamp): """ self.__logs.append((id, timestamp)) - + def retrieve(self, s, e, gra): """ :type s: str diff --git a/Python/design-phone-directory.py b/Python/design-phone-directory.py index d0c2af2a1..2d039ca18 100644 --- a/Python/design-phone-directory.py +++ b/Python/design-phone-directory.py @@ -28,7 +28,7 @@ def get(self): self.__curr += 1 self.__used[number] = True return number - + def check(self, number): """ @@ -38,7 +38,7 @@ def check(self, number): """ return 0 <= number < len(self.__numbers) and \ not self.__used[number] - + def release(self, number): """ @@ -52,7 +52,7 @@ def release(self, number): self.__used[number] = False self.__curr -= 1 self.__numbers[self.__curr] = number - + # Your PhoneDirectory object will be instantiated and called as such: # obj = PhoneDirectory(maxNumbers) diff --git a/Python/diagonal-traverse.py b/Python/diagonal-traverse.py index 8410476a4..639c64adf 100644 --- a/Python/diagonal-traverse.py +++ b/Python/diagonal-traverse.py @@ -26,11 +26,11 @@ def findDiagonalOrder(self, matrix): """ if not matrix or not matrix[0]: return [] - + result = [] row, col, d = 0, 0, 0 dirs = [(-1, 1), (1, -1)] - + for i in xrange(len(matrix) * len(matrix[0])): result.append(matrix[row][col]) row += dirs[d][0] @@ -50,6 +50,6 @@ def findDiagonalOrder(self, matrix): elif col < 0: col = 0 d = 1 - d - + return result - + diff --git a/Python/diameter-of-binary-tree.py b/Python/diameter-of-binary-tree.py index 736e93b3b..063a60ffc 100644 --- a/Python/diameter-of-binary-tree.py +++ b/Python/diameter-of-binary-tree.py @@ -6,12 +6,12 @@ # any two nodes in a tree. This path may or may not pass through the root. # # Example: -# Given a binary tree +# Given a binary tree # 1 # / \ # 2 3 -# / \ -# 4 5 +# / \ +# 4 5 # Return 3, which is the length of the path [4,2,1,3] or [5,2,1,3]. # # Note: The length of path between two nodes is represented by the number of edges between them. @@ -34,5 +34,5 @@ def depth(root, diameter): left, diameter = depth(root.left, diameter) right, diameter = depth(root.right, diameter) return 1 + max(left, right), max(diameter, 1 + left + right) - + return depth(root, 1)[1] - 1 diff --git a/Python/distinct-subsequences.py b/Python/distinct-subsequences.py index 48709899e..5d4aad7bc 100644 --- a/Python/distinct-subsequences.py +++ b/Python/distinct-subsequences.py @@ -2,14 +2,14 @@ # Space: O(n) # # Given a string S and a string T, count the number of distinct subsequences of T in S. -# -# A subsequence of a string is a new string which is formed from the original string -# by deleting some (can be none) of the characters without disturbing the relative positions +# +# A subsequence of a string is a new string which is formed from the original string +# by deleting some (can be none) of the characters without disturbing the relative positions # of the remaining characters. (ie, "ACE" is a subsequence of "ABCDE" while "AEC" is not). -# +# # Here is an example: # S = "rabbbit", T = "rabbit" -# +# # Return 3. # @@ -23,10 +23,10 @@ def numDistinct(self, S, T): if S_char == T_char: ways[j + 1] += ways[j] return ways[len(T)] - + if __name__ == "__main__": S = "rabbbit" T = "rabbit" result = Solution().numDistinct(S, T) print result - + diff --git a/Python/distribute-candies.py b/Python/distribute-candies.py index a04064883..827919c11 100644 --- a/Python/distribute-candies.py +++ b/Python/distribute-candies.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(n) -# Given an integer array with even length, where different numbers +# Given an integer array with even length, where different numbers # in this array represent different kinds of candies. # Each number means one candy of the corresponding kind. # You need to distribute these candies equally in number to brother and sister. @@ -12,14 +12,14 @@ # Output: 3 # Explanation: # There are three different kinds of candies (1, 2 and 3), and two candies for each kind. -# Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too. -# The sister has three different kinds of candies. +# Optimal distribution: The sister has candies [1,2,3] and the brother has candies [1,2,3], too. +# The sister has three different kinds of candies. # # Example 2: # Input: candies = [1,1,2,3] # Output: 2 -# Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1]. -# The sister has two different kinds of candies, the brother has only one kind of candies. +# Explanation: For example, the sister has candies [2,3] and the brother has candies [1,1]. +# The sister has two different kinds of candies, the brother has only one kind of candies. # # Note: # The length of the given array is in range [2, 10,000], and will be even. diff --git a/Python/divide-two-integers.py b/Python/divide-two-integers.py index b024170f6..9e2bd8193 100644 --- a/Python/divide-two-integers.py +++ b/Python/divide-two-integers.py @@ -1,6 +1,6 @@ # Time: O(logn) = O(1) # Space: O(1) -# +# # Divide two integers without using multiplication, division and mod operator. @@ -44,7 +44,7 @@ def divide2(self, dividend, divisor): if not positive: res = -res return min(max(-2147483648, res), 2147483647) - + if __name__ == "__main__": print Solution().divide(123, 12) print Solution().divide(123, -12) diff --git a/Python/dungeon-game.py b/Python/dungeon-game.py index 9348828cb..d42e5d1a3 100644 --- a/Python/dungeon-game.py +++ b/Python/dungeon-game.py @@ -1,33 +1,33 @@ # Time: O(m * n) # Space: O(m + n) # -# The demons had captured the princess (P) and imprisoned her +# The demons had captured the princess (P) and imprisoned her # in the bottom-right corner of a dungeon. T -# he dungeon consists of M x N rooms laid out in a 2D grid. -# Our valiant knight (K) was initially positioned in the top-left room +# he dungeon consists of M x N rooms laid out in a 2D grid. +# Our valiant knight (K) was initially positioned in the top-left room # and must fight his way through the dungeon to rescue the princess. -# -# The knight has an initial health point represented by a positive integer. +# +# The knight has an initial health point represented by a positive integer. # If at any point his health point drops to 0 or below, he dies immediately. -# -# Some of the rooms are guarded by demons, -# so the knight loses health (negative integers) upon entering these rooms; +# +# Some of the rooms are guarded by demons, +# so the knight loses health (negative integers) upon entering these rooms; # other rooms are either empty (0's) or contain magic orbs that increase the knight's health (positive integers). -# -# In order to reach the princess as quickly as possible, +# +# In order to reach the princess as quickly as possible, # the knight decides to move only rightward or downward in each step. -# -# -# Write a function to determine the knight's minimum initial health +# +# +# Write a function to determine the knight's minimum initial health # so that he is able to rescue the princess. -# -# For example, given the dungeon below, the initial health of +# +# For example, given the dungeon below, the initial health of # the knight must be at least 7 if he follows the optimal path RIGHT-> RIGHT -> DOWN -> DOWN. -# +# # Notes: -# +# # The knight's health has no upper bound. -# Any room can contain threats or power-ups, even the first room the knight enters +# Any room can contain threats or power-ups, even the first room the knight enters # and the bottom-right room where the princess is imprisoned. # @@ -37,16 +37,16 @@ class Solution: def calculateMinimumHP(self, dungeon): DP = [float("inf") for _ in dungeon[0]] DP[-1] = 1 - + for i in reversed(xrange(len(dungeon))): DP[-1] = max(DP[-1] - dungeon[i][-1], 1) for j in reversed(xrange(len(dungeon[i]) - 1)): min_HP_on_exit = min(DP[j], DP[j + 1]) DP[j] = max(min_HP_on_exit - dungeon[i][j], 1) - + return DP[0] -# Time: O(m * n logk), where k is the possible maximum sum of loses +# Time: O(m * n logk), where k is the possible maximum sum of loses # Space: O(m + n) class Solution2: # @param dungeon, a list of lists of integers @@ -57,9 +57,9 @@ def calculateMinimumHP(self, dungeon): for room in rooms: if room < 0: maximum_loses += abs(room) - + return self.binarySearch(dungeon, maximum_loses) - + def binarySearch(self, dungeon, maximum_loses): start, end = 1, maximum_loses + 1 result = 0 @@ -70,20 +70,20 @@ def binarySearch(self, dungeon, maximum_loses): else: start = mid + 1 return start - + def DP(self, dungeon, HP): remain_HP = [0 for _ in dungeon[0]] remain_HP[0] = HP + dungeon[0][0] for j in xrange(1, len(remain_HP)): - if remain_HP[j - 1] > 0: + if remain_HP[j - 1] > 0: remain_HP[j] = max(remain_HP[j - 1] + dungeon[0][j], 0) - + for i in xrange(1, len(dungeon)): if remain_HP[0] > 0: remain_HP[0] = max(remain_HP[0] + dungeon[i][0], 0) else: remain_HP[0] = 0 - + for j in xrange(1, len(remain_HP)): remain = 0 if remain_HP[j - 1] > 0: @@ -99,9 +99,9 @@ def DP(self, dungeon, HP): [ -5, -10, 1], \ [ 10, 30, -5]] print Solution().calculateMinimumHP(dungeon) - + dungeon = [[ -200]] print Solution().calculateMinimumHP(dungeon) - + dungeon = [[0, -3]] print Solution().calculateMinimumHP(dungeon) \ No newline at end of file diff --git a/Python/edit-distance.py b/Python/edit-distance.py index 8498a2f57..2ca1e67c4 100644 --- a/Python/edit-distance.py +++ b/Python/edit-distance.py @@ -1,11 +1,11 @@ # Time: O(n * m) # Space: O(n + m) # -# Given two words word1 and word2, find the minimum number of steps +# Given two words word1 and word2, find the minimum number of steps # required to convert word1 to word2. (each operation is counted as 1 step.) -# +# # You have the following 3 operations permitted on a word: -# +# # a) Insert a character # b) Delete a character # c) Replace a character @@ -16,9 +16,9 @@ class Solution: def minDistance(self, word1, word2): if len(word1) < len(word2): return self.minDistance(word2, word1) - + distance = [i for i in xrange(len(word2) + 1)] - + for i in xrange(1, len(word1) + 1): pre_distance_i_j = distance[0] distance[0] = i @@ -37,10 +37,10 @@ def minDistance(self, word1, word2): # Space: O(n * m) class Solution2: # @return an integer - def minDistance(self, word1, word2): + def minDistance(self, word1, word2): distance = [[i] for i in xrange(len(word1) + 1)] distance[0] = [j for j in xrange(len(word2) + 1)] - + for i in xrange(1, len(word1) + 1): for j in xrange(1, len(word2) + 1): insert = distance[i][j - 1] + 1 @@ -49,7 +49,7 @@ def minDistance(self, word1, word2): if word1[i - 1] != word2[j - 1]: replace += 1 distance[i].append(min(insert, delete, replace)) - + return distance[-1][-1] if __name__ == "__main__": diff --git a/Python/encode-and-decode-strings.py b/Python/encode-and-decode-strings.py index 39445a51a..b6e33bbda 100644 --- a/Python/encode-and-decode-strings.py +++ b/Python/encode-and-decode-strings.py @@ -5,7 +5,7 @@ class Codec: def encode(self, strs): """Encodes a list of strings to a single string. - + :type strs: List[str] :rtype: str """ @@ -17,7 +17,7 @@ def encode(self, strs): def decode(self, s): """Decodes a single string to a list of strings. - + :type s: str :rtype: List[str] """ @@ -26,5 +26,5 @@ def decode(self, s): while i < len(s): l = int(s[i:i+8], 16) strs.append(s[i+8:i+8+l]) - i += 8+l + i += 8+l return strs diff --git a/Python/encode-string-with-shortest-length.py b/Python/encode-string-with-shortest-length.py index 6988e6d4c..667663b29 100644 --- a/Python/encode-string-with-shortest-length.py +++ b/Python/encode-string-with-shortest-length.py @@ -23,6 +23,6 @@ def encode_substr(dp, s, i, j): if len(dp[i][k]) + len(dp[k+1][j]) < len(dp[i][j]): dp[i][j] = dp[i][k] + dp[k+1][j] encoded_string = encode_substr(dp, s, i, j) - if len(encoded_string) < len(dp[i][j]): + if len(encoded_string) < len(dp[i][j]): dp[i][j] = encoded_string return dp[0][len(s) - 1] diff --git a/Python/escape-the-ghosts.py b/Python/escape-the-ghosts.py index be1e5f43f..0dff9431f 100644 --- a/Python/escape-the-ghosts.py +++ b/Python/escape-the-ghosts.py @@ -17,27 +17,27 @@ # Return True if and only if it is possible to escape. # # Example 1: -# Input: +# Input: # ghosts = [[1, 0], [0, 3]] # target = [0, 1] # Output: true -# Explanation: +# Explanation: # You can directly reach the destination (0, 1) at time 1, # while the ghosts located at (1, 0) or (0, 3) have no way to catch up with you. # # Example 2: -# Input: +# Input: # ghosts = [[1, 0]] # target = [2, 0] # Output: false -# Explanation: +# Explanation: # You need to reach the destination (2, 0), but the ghost at (1, 0) lies between you and the destination. # Example 3: -# Input: +# Input: # ghosts = [[2, 0]] # target = [1, 0] # Output: false -# Explanation: +# Explanation: # The ghost can reach the target at the same time as you. # # Note: diff --git a/Python/evaluate-reverse-polish-notation.py b/Python/evaluate-reverse-polish-notation.py index e437aaf09..b64bad971 100644 --- a/Python/evaluate-reverse-polish-notation.py +++ b/Python/evaluate-reverse-polish-notation.py @@ -2,9 +2,9 @@ # Space: O(n) # # Evaluate the value of an arithmetic expression in Reverse Polish Notation. -# +# # Valid operators are +, -, *, /. Each operand may be an integer or another expression. -# +# # Some examples: # ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9 # ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6 @@ -23,7 +23,7 @@ def evalRPN(self, tokens): y, x = numerals.pop(), numerals.pop() numerals.append(int(operators[token](x * 1.0, y))) return numerals.pop() - + if __name__ == "__main__": print Solution().evalRPN(["2", "1", "+", "3", "*"]) print Solution().evalRPN(["4", "13", "5", "/", "+"]) diff --git a/Python/excel-sheet-column-number.py b/Python/excel-sheet-column-number.py index 667233c59..ad16cb326 100644 --- a/Python/excel-sheet-column-number.py +++ b/Python/excel-sheet-column-number.py @@ -2,18 +2,18 @@ # Space: O(1) # Related to question Excel Sheet Column Title -# +# # Given a column title as appear in an Excel sheet, return its corresponding column number. -# +# # For example: -# +# # A -> 1 # B -> 2 # C -> 3 # ... # Z -> 26 # AA -> 27 -# AB -> 28 +# AB -> 28 class Solution(object): def titleToNumber(self, s): diff --git a/Python/excel-sheet-column-title.py b/Python/excel-sheet-column-title.py index a2160e9d5..041031253 100644 --- a/Python/excel-sheet-column-title.py +++ b/Python/excel-sheet-column-title.py @@ -2,16 +2,16 @@ # Space: O(1) # Given a positive integer, return its corresponding column title as appear in an Excel sheet. -# +# # For example: -# +# # 1 -> A # 2 -> B # 3 -> C # ... # 26 -> Z # 27 -> AA -# 28 -> AB +# 28 -> AB class Solution(object): def convertToTitle(self, n): @@ -20,11 +20,11 @@ def convertToTitle(self, n): :rtype: str """ result, dvd = "", n - + while dvd: result += chr((dvd - 1) % 26 + ord('A')) dvd = (dvd - 1) / 26 - + return result[::-1] diff --git a/Python/exclusive-time-of-functions.py b/Python/exclusive-time-of-functions.py index 02127e25c..fff0bcbf7 100644 --- a/Python/exclusive-time-of-functions.py +++ b/Python/exclusive-time-of-functions.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(n) -# Given the running logs of n functions that are executed +# Given the running logs of n functions that are executed # in a nonpreemptive single threaded CPU, find the exclusive time of these functions. # # Each function has a unique id, start from 0 to n-1. @@ -19,7 +19,7 @@ # Example 1: # Input: # n = 2 -# logs = +# logs = # ["0:start:0", # "1:start:2", # "1:end:5", @@ -27,9 +27,9 @@ # Output:[3, 4] # # Explanation: -# Function 0 starts at time 0, then it executes 2 units of time and reaches the end of time 1. +# Function 0 starts at time 0, then it executes 2 units of time and reaches the end of time 1. # Now function 0 calls function 1, function 1 starts at time 2, executes 4 units of time and end at time 5. -# Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. +# Function 0 is running again at time 6, and also end at the time 6, thus executes 1 unit of time. # So function 0 totally execute 2 + 1 = 3 units of time, and function 1 totally execute 4 units of time. # # Note: diff --git a/Python/expression-add-operators.py b/Python/expression-add-operators.py index 15d972d3f..f26c4c7ad 100644 --- a/Python/expression-add-operators.py +++ b/Python/expression-add-operators.py @@ -1,13 +1,13 @@ # Time: O(4^n) # Space: O(n) # -# Given a string that contains only digits 0-9 -# and a target value, return all possibilities +# Given a string that contains only digits 0-9 +# and a target value, return all possibilities # to add operators +, -, or * between the digits # so they evaluate to the target value. # -# Examples: -# "123", 6 -> ["1+2+3", "1*2*3"] +# Examples: +# "123", 6 -> ["1+2+3", "1*2*3"] # "232", 8 -> ["2*3+2", "2+3*2"] # "00", 0 -> ["0+0", "0-0", "0*0"] # "3456237490", 9191 -> [] @@ -47,21 +47,21 @@ def addOperatorsDFS(self, num, target, pos, operand1, operand2, expr, result): # Avoid "00...". if str(val) != val_str: break - + # Case '+': expr.append("+" + val_str) self.addOperatorsDFS(num, target, i + 1, operand1 + operand2, val, expr, result) expr.pop() - + # Case '-': expr.append("-" + val_str) self.addOperatorsDFS(num, target, i + 1, operand1 + operand2, -val, expr, result) expr.pop() - + # Case '*': expr.append("*" + val_str) self.addOperatorsDFS(num, target, i + 1, operand1, operand2 * val, expr, result) expr.pop() - + i += 1 - + diff --git a/Python/factorial-trailing-zeroes.py b/Python/factorial-trailing-zeroes.py index 106b59beb..bccfd7e75 100644 --- a/Python/factorial-trailing-zeroes.py +++ b/Python/factorial-trailing-zeroes.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given an integer n, return the number of trailing zeroes in n!. -# +# # Note: Your solution should be in logarithmic time complexity. # diff --git a/Python/find-all-anagrams-in-a-string.py b/Python/find-all-anagrams-in-a-string.py index e80e46ee0..4122bc0bc 100644 --- a/Python/find-all-anagrams-in-a-string.py +++ b/Python/find-all-anagrams-in-a-string.py @@ -45,7 +45,7 @@ def findAnagrams(self, s, p): cnts = [0] * 26 for c in p: cnts[ord(c) - ord('a')] += 1 - + left, right = 0, 0 while right < len(s): cnts[ord(s[right]) - ord('a')] -= 1 diff --git a/Python/find-all-duplicates-in-an-array.py b/Python/find-all-duplicates-in-an-array.py index a9f4f054d..47a601d3a 100644 --- a/Python/find-all-duplicates-in-an-array.py +++ b/Python/find-all-duplicates-in-an-array.py @@ -28,7 +28,7 @@ def findDuplicates(self, nums): else: nums[abs(i)-1] *= -1 return result - + # Time: O(n) # Space: O(1) @@ -42,7 +42,7 @@ def findDuplicates(self, nums): i = 0 while i < len(nums): if nums[i] != nums[nums[i]-1]: - nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1] + nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1] else: i += 1 diff --git a/Python/find-bottom-left-tree-value.py b/Python/find-bottom-left-tree-value.py index bda1f8710..15dbde147 100644 --- a/Python/find-bottom-left-tree-value.py +++ b/Python/find-bottom-left-tree-value.py @@ -12,7 +12,7 @@ # # Output: # 1 -# Example 2: +# Example 2: # Input: # # 1 diff --git a/Python/find-k-pairs-with-smallest-sums.py b/Python/find-k-pairs-with-smallest-sums.py index 272a4ccff..6a1750aee 100644 --- a/Python/find-k-pairs-with-smallest-sums.py +++ b/Python/find-k-pairs-with-smallest-sums.py @@ -24,7 +24,7 @@ # The first 2 pairs are returned from the sequence: # [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3] # Example 3: -# Given nums1 = [1,2], nums2 = [3], k = 3 +# Given nums1 = [1,2], nums2 = [3], k = 3 # # Return: [1,3],[2,3] # diff --git a/Python/find-k-th-smallest-pair-distance.py b/Python/find-k-th-smallest-pair-distance.py index 6badaf3c1..970fe106a 100644 --- a/Python/find-k-th-smallest-pair-distance.py +++ b/Python/find-k-th-smallest-pair-distance.py @@ -8,7 +8,7 @@ # Input: # nums = [1,3,1] # k = 1 -# Output: 0 +# Output: 0 # Explanation: # Here are all the pairs: # (1,3) -> 2 diff --git a/Python/find-largest-value-in-each-tree-row.py b/Python/find-largest-value-in-each-tree-row.py index 0ddc4af54..0343ca2a7 100644 --- a/Python/find-largest-value-in-each-tree-row.py +++ b/Python/find-largest-value-in-each-tree-row.py @@ -4,13 +4,13 @@ # You need to find the largest value in each row of a binary tree. # # Example: -# Input: +# Input: # # 1 # / \ # 3 2 -# / \ \ -# 5 3 9 +# / \ \ +# 5 3 9 # # Output: [1, 3, 9] @@ -56,4 +56,4 @@ def largestValues(self, root): result.append(max(node.val for node in curr)) curr = [child for node in curr for child in (node.left, node.right) if child] return result - + diff --git a/Python/find-median-from-data-stream.py b/Python/find-median-from-data-stream.py index 3e3f968fb..df79d898e 100644 --- a/Python/find-median-from-data-stream.py +++ b/Python/find-median-from-data-stream.py @@ -1,11 +1,11 @@ # Time: O(nlogn) for total n addNums, O(logn) per addNum, O(1) per findMedian. # Space: O(n), total space -# Median is the middle value in an ordered integer list. -# If the size of the list is even, there is no middle value. +# Median is the middle value in an ordered integer list. +# If the size of the list is even, there is no middle value. # So the median is the mean of the two middle value. # -# Examples: +# Examples: # [2,3,4] , the median is 3 # # [2,3], the median is (2 + 3) / 2 = 2.5 @@ -19,7 +19,7 @@ # add(1) # add(2) # findMedian() -> 1.5 -# add(3) +# add(3) # findMedian() -> 2 # Heap solution. diff --git a/Python/find-minimum-in-rotated-sorted-array-ii.py b/Python/find-minimum-in-rotated-sorted-array-ii.py index fcb09d212..911eacbb7 100644 --- a/Python/find-minimum-in-rotated-sorted-array-ii.py +++ b/Python/find-minimum-in-rotated-sorted-array-ii.py @@ -3,14 +3,14 @@ # # Follow up for "Find Minimum in Rotated Sorted Array": # What if duplicates are allowed? -# +# # Would this affect the run-time complexity? How and why? # Suppose a sorted array is rotated at some pivot unknown to you beforehand. -# +# # (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). -# +# # Find the minimum element. -# +# # The array may contain duplicates. # @@ -43,7 +43,7 @@ def findMin(self, nums): left, right = 0, len(nums) - 1 while left < right and nums[left] >= nums[right]: mid = left + (right - left) / 2 - + if nums[mid] == nums[left]: left += 1 elif nums[mid] < nums[left]: diff --git a/Python/find-mode-in-binary-search-tree.py b/Python/find-mode-in-binary-search-tree.py index 60a53fa65..facb55471 100644 --- a/Python/find-mode-in-binary-search-tree.py +++ b/Python/find-mode-in-binary-search-tree.py @@ -39,7 +39,7 @@ def findMode(self, root): def inorder(root, prev, cnt, max_cnt, result): if not root: return prev, cnt, max_cnt - + prev, cnt, max_cnt = inorder(root.left, prev, cnt, max_cnt, result) if prev: if root.val == prev.val: @@ -53,7 +53,7 @@ def inorder(root, prev, cnt, max_cnt, result): elif cnt == max_cnt: result.append(root.val) return inorder(root.right, root, cnt, max_cnt, result) - + if not root: return [] result = [] diff --git a/Python/find-peak-element.py b/Python/find-peak-element.py index dfa85fd2d..b986c43bf 100644 --- a/Python/find-peak-element.py +++ b/Python/find-peak-element.py @@ -2,15 +2,15 @@ # Space: O(1) # A peak element is an element that is greater than its neighbors. -# +# # Given an input array where num[i] != num[i+1], # find a peak element and return its index. -# -# The array may contain multiple peaks, in that case +# +# The array may contain multiple peaks, in that case # return the index to any one of the peaks is fine. -# +# # You may imagine that num[-1] = num[n] = -infinite. -# +# # For example, in array [1, 2, 3, 1], 3 is a peak element # and your function should return the index number 2. # @@ -18,21 +18,21 @@ # Your solution should be in logarithmic complexity. -class Solution(object): +class Solution(object): def findPeakElement(self, nums): """ :type nums: List[int] :rtype: int """ left, right = 0, len(nums) - 1 - + while left < right: mid = left + (right - left) / 2 if nums[mid] > nums[mid + 1]: right = mid else: left = mid + 1 - + return left diff --git a/Python/find-pivot-index.py b/Python/find-pivot-index.py index 5ca035a46..37880950f 100644 --- a/Python/find-pivot-index.py +++ b/Python/find-pivot-index.py @@ -10,18 +10,18 @@ # you should return the left-most pivot index. # # Example 1: -# Input: +# Input: # nums = [1, 7, 3, 6, 5, 6] # Output: 3 -# Explanation: +# Explanation: # The sum of the numbers to the left of index 3 (nums[3] = 6) is equal to the sum of numbers to the right of index 3. # Also, 3 is the first index where this occurs. # # Example 2: -# Input: +# Input: # nums = [1, 2, 3] # Output: -1 -# Explanation: +# Explanation: # There is no index that satisfies the conditions in the problem statement. # # Note: @@ -41,4 +41,4 @@ def pivotIndex(self, nums): return i left_sum += num return -1 - + diff --git a/Python/find-the-duplicate-number.py b/Python/find-the-duplicate-number.py index c7c9fe49d..011afc5be 100644 --- a/Python/find-the-duplicate-number.py +++ b/Python/find-the-duplicate-number.py @@ -55,7 +55,7 @@ def findDuplicate(self, nums): if num <= mid: count += 1 if count > mid: - right = mid - 1 + right = mid - 1 else: left = mid + 1 return left diff --git a/Python/first-bad-version.py b/Python/first-bad-version.py index af2bdc0a5..32ff42d2b 100644 --- a/Python/first-bad-version.py +++ b/Python/first-bad-version.py @@ -4,16 +4,16 @@ # You are a product manager and currently leading a team to # develop a new product. Unfortunately, the latest version of # your product fails the quality check. Since each version is -# developed based on the previous version, all the versions +# developed based on the previous version, all the versions # after a bad version are also bad. # -# Suppose you have n versions [1, 2, ..., n] and you want to -# find out the first bad one, which causes all the following +# Suppose you have n versions [1, 2, ..., n] and you want to +# find out the first bad one, which causes all the following # ones to be bad. # -# You are given an API bool isBadVersion(version) which will -# return whether version is bad. Implement a function to find -# the first bad version. You should minimize the number of +# You are given an API bool isBadVersion(version) which will +# return whether version is bad. Implement a function to find +# the first bad version. You should minimize the number of # calls to the API. # diff --git a/Python/first-missing-positive.py b/Python/first-missing-positive.py index 134baa714..eb09936d4 100644 --- a/Python/first-missing-positive.py +++ b/Python/first-missing-positive.py @@ -2,11 +2,11 @@ # Space: O(1) # # Given an unsorted integer array, find the first missing positive integer. -# +# # For example, # Given [1,2,0] return 3, # and [3,4,-1,1] return 2. -# +# # Your algorithm should run in O(n) time and uses constant space. # @@ -20,12 +20,12 @@ def firstMissingPositive(self, A): A[A[i]-1], A[i] = A[i], A[A[i]-1] else: i += 1 - + for i, integer in enumerate(A): if integer != i + 1: return i + 1 return len(A) + 1 - + if __name__ == "__main__": print Solution().firstMissingPositive([1,2,0]) print Solution().firstMissingPositive([3,4,-1,1]) diff --git a/Python/flatten-2d-vector.py b/Python/flatten-2d-vector.py index 7b40db259..7443014cb 100644 --- a/Python/flatten-2d-vector.py +++ b/Python/flatten-2d-vector.py @@ -26,7 +26,7 @@ def hasNext(self): return self.x != len(self.vec) and self.y != len(self.vec[self.x]) def adjustNextIter(self): - while self.x != len(self.vec) and self.y == len(self.vec[self.x]): + while self.x != len(self.vec) and self.y == len(self.vec[self.x]): self.x += 1 if self.x != len(self.vec): self.y = 0 diff --git a/Python/flatten-binary-tree-to-linked-list.py b/Python/flatten-binary-tree-to-linked-list.py index 579f2e319..581275cb0 100644 --- a/Python/flatten-binary-tree-to-linked-list.py +++ b/Python/flatten-binary-tree-to-linked-list.py @@ -2,10 +2,10 @@ # Space: O(h), h is height of binary tree # # Given a binary tree, flatten it to a linked list in-place. -# +# # For example, # Given -# +# # 1 # / \ # 2 5 @@ -37,7 +37,7 @@ class Solution: # @return nothing, do it in place def flatten(self, root): return self.flattenRecu(root, None) - + def flattenRecu(self, root, list_head): if root != None: list_head = self.flattenRecu(root.right, list_head) @@ -47,7 +47,7 @@ def flattenRecu(self, root, list_head): return root else: return list_head - + class Solution2: list_head = None # @param root, a tree node @@ -60,7 +60,7 @@ def flatten(self, root): root.left = None self.list_head = root return root - + if __name__ == "__main__": root = TreeNode(1) root.left = TreeNode(2) diff --git a/Python/flatten-nested-list-iterator.py b/Python/flatten-nested-list-iterator.py index e9646fb69..255847377 100644 --- a/Python/flatten-nested-list-iterator.py +++ b/Python/flatten-nested-list-iterator.py @@ -59,7 +59,7 @@ def hasNext(self): self.__depth[-1][1] += 1 self.__depth.append([nestedList[i].getList(), 0]) return False - + # Your NestedIterator object will be instantiated and called as such: # i, v = NestedIterator(nestedList), [] diff --git a/Python/flip-game.py b/Python/flip-game.py index 8c8760f43..b366ed982 100644 --- a/Python/flip-game.py +++ b/Python/flip-game.py @@ -29,4 +29,4 @@ def generatePossibleNextMoves(self, s): :rtype: List[str] """ return [s[:i] + "--" + s[i+2:] for i in xrange(len(s) - 1) if s[i:i+2] == "++"] - + diff --git a/Python/flood-fill.py b/Python/flood-fill.py index 7f1149a0f..701117bea 100644 --- a/Python/flood-fill.py +++ b/Python/flood-fill.py @@ -15,12 +15,12 @@ # At the end, return the modified image. # # Example 1: -# Input: +# Input: # image = [[1,1,1],[1,1,0],[1,0,1]] # sr = 1, sc = 1, newColor = 2 # Output: [[2,2,2],[2,2,0],[2,0,1]] -# Explanation: -# From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected +# Explanation: +# From the center of the image (with position (sr, sc) = (1, 1)), all pixels connected # by a path of the same color as the starting pixel are colored with the new color. # Note the bottom corner is not colored 2, because it is not 4-directionally connected # to the starting pixel. @@ -46,11 +46,11 @@ def dfs(image, r, c, newColor, color): 0 <= c < len(image[0]) and \ image[r][c] == color): return - + image[r][c] = newColor for d in directions: dfs(image, r+d[0], c+d[1], newColor, color) - + color = image[sr][sc] if color == newColor: return image dfs(image, sr, sc, newColor, color) diff --git a/Python/fraction-to-recurring-decimal.py b/Python/fraction-to-recurring-decimal.py index 98dff2202..7fed92fb2 100644 --- a/Python/fraction-to-recurring-decimal.py +++ b/Python/fraction-to-recurring-decimal.py @@ -3,11 +3,11 @@ # Given two integers representing the numerator and denominator of a fraction, # return the fraction in string format. -# +# # If the fractional part is repeating, enclose the repeating part in parentheses. -# +# # For example, -# +# # Given numerator = 1, denominator = 2, return "0.5". # Given numerator = 2, denominator = 1, return "2". # Given numerator = 2, denominator = 3, return "0.(6)". @@ -20,8 +20,8 @@ def fractionToDecimal(self, numerator, denominator): :rtype: str """ result = "" - if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0): - result = "-" + if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0): + result = "-" dvd, dvs = abs(numerator), abs(denominator) result += str(dvd / dvs) diff --git a/Python/friend-circles.py b/Python/friend-circles.py index 19ee1f376..486294dfd 100644 --- a/Python/friend-circles.py +++ b/Python/friend-circles.py @@ -12,21 +12,21 @@ # otherwise not. And you have to output the total number of friend circles among all the students. # # Example 1: -# Input: +# Input: # [[1,1,0], # [1,1,0], # [0,0,1]] # Output: 2 -# Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. +# Explanation:The 0th and 1st students are direct friends, so they are in a friend circle. # The 2nd student himself is in a friend circle. So return 2. # # Example 2: -# Input: +# Input: # [[1,1,0], # [1,1,1], # [0,1,1]] # Output: 1 -# Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, +# Explanation:The 0th and 1st students are direct friends, the 1st and 2nd students are direct friends, # so the 0th and 2nd students are indirect friends. All of them are in the same friend circle, so return 1. # # Note: @@ -45,12 +45,12 @@ class UnionFind(object): def __init__(self, n): self.set = range(n) self.count = n - + def find_set(self, x): if self.set[x] != x: self.set[x] = self.find_set(self.set[x]) # path compression. return self.set[x] - + def union_set(self, x, y): x_root, y_root = map(self.find_set, (x, y)) if x_root != y_root: @@ -63,4 +63,4 @@ def union_set(self, x, y): if M[i][j] and i != j: circles.union_set(i, j) return circles.count - + diff --git a/Python/frog-jump.py b/Python/frog-jump.py index 447071fbe..3b1ca8637 100644 --- a/Python/frog-jump.py +++ b/Python/frog-jump.py @@ -1,8 +1,8 @@ -# Time: O(n) ~ O(n^2) +# Time: O(n) ~ O(n^2) # Space: O(n) # A frog is crossing a river. The river is divided into x units and -# at each unit there may or may not exist a stone. +# at each unit there may or may not exist a stone. # The frog can jump on a stone, but it must not jump into the water. # # Given a list of stones' positions (in units) in sorted ascending order, @@ -26,15 +26,15 @@ # third stone at the 3rd unit, and so on... # The last stone at the 17th unit. # -# Return true. The frog can jump to the last stone by jumping -# 1 unit to the 2nd stone, then 2 units to the 3rd stone, then -# 2 units to the 4th stone, then 3 units to the 6th stone, +# Return true. The frog can jump to the last stone by jumping +# 1 unit to the 2nd stone, then 2 units to the 3rd stone, then +# 2 units to the 4th stone, then 3 units to the 6th stone, # 4 units to the 7th stone, and 5 units to the 8th stone. # Example 2: # # [0,1,2,3,4,8,9,11] # -# Return false. There is no way to jump to the last stone as +# Return false. There is no way to jump to the last stone as # the gap between the 5th and 6th stone is too large. # DP with hash table diff --git a/Python/game-of-life.py b/Python/game-of-life.py index f569bde1f..8e2a4132a 100644 --- a/Python/game-of-life.py +++ b/Python/game-of-life.py @@ -1,35 +1,35 @@ # Time: O(m * n) # Space: O(1) -# According to the Wikipedia's article: -# "The Game of Life, also known simply as Life, -# is a cellular automaton devised by the British +# According to the Wikipedia's article: +# "The Game of Life, also known simply as Life, +# is a cellular automaton devised by the British # mathematician John Horton Conway in 1970." # -# Given a board with m by n cells, each cell has -# an initial state live (1) or dead (0). +# Given a board with m by n cells, each cell has +# an initial state live (1) or dead (0). # Each cell interacts with its eight neighbors # (horizontal, vertical, diagonal) -# using the following four rules +# using the following four rules # (taken from the above Wikipedia article): # # - Any live cell with fewer than two live neighbors dies, # as if caused by under-population. -# - Any live cell with two or three live neighbors lives +# - Any live cell with two or three live neighbors lives # on to the next generation. # - Any live cell with more than three live neighbors dies, # as if by over-population.. -# - Any dead cell with exactly three live neighbors +# - Any dead cell with exactly three live neighbors # becomes a live cell, as if by reproduction. # -# Write a function to compute the next state +# Write a function to compute the next state # (after one update) of the board given its current state. # -# Follow up: +# Follow up: # - Could you solve it in-place? Remember that the board needs # to be updated at the same time: You cannot update some cells # first and then use their updated values to update other cells. -# - In this question, we represent the board using a 2D array. +# - In this question, we represent the board using a 2D array. # In principle, the board is infinite, which would cause problems # when the active area encroaches the border of the array. # How would you address these problems? @@ -57,7 +57,7 @@ def gameOfLife(self, board): # Any live cell with two live neighbors. # Any dead cell with exactly three live neighbors lives. if (count == 4 and board[i][j]) or count == 3: - board[i][j] |= 2 # Mark as live. + board[i][j] |= 2 # Mark as live. for i in xrange(m): for j in xrange(n): diff --git a/Python/gas-station.py b/Python/gas-station.py index 72d6a0806..30a6fe5cc 100644 --- a/Python/gas-station.py +++ b/Python/gas-station.py @@ -2,12 +2,12 @@ # Space: O(1) # # There are N gas stations along a circular route, where the amount of gas at station i is gas[i]. -# +# # You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). # You begin the journey with an empty tank at one of the gas stations. -# +# # Return the starting gas station's index if you can travel around the circuit once, otherwise return -1. -# +# # Note: # The solution is guaranteed to be unique. # @@ -27,7 +27,7 @@ def canCompleteCircuit(self, gas, cost): current_sum = 0 if total_sum >= 0: return start - + return -1 if __name__ == "__main__": diff --git a/Python/generate-parentheses.py b/Python/generate-parentheses.py index 0b9712875..2ffcfc194 100644 --- a/Python/generate-parentheses.py +++ b/Python/generate-parentheses.py @@ -1,11 +1,11 @@ # Time: O(4^n / n^(3/2)) ~= Catalan numbers # Space: O(n) # -# Given n pairs of parentheses, write a function to generate +# Given n pairs of parentheses, write a function to generate # all combinations of well-formed parentheses. -# +# # For example, given n = 3, a solution set is: -# +# # "((()))", "(()())", "(())()", "()(())", "()()()" # @@ -16,7 +16,7 @@ def generateParenthesis(self, n): result = [] self.generateParenthesisRecu(result, "", n, n) return result - + def generateParenthesisRecu(self, result, current, left, right): if left == 0 and right == 0: result.append(current) diff --git a/Python/global-and-local-inversions.py b/Python/global-and-local-inversions.py index 2ee8525b0..2c7e22024 100644 --- a/Python/global-and-local-inversions.py +++ b/Python/global-and-local-inversions.py @@ -30,4 +30,4 @@ def isIdealPermutation(self, A): :rtype: bool """ return all(abs(v-i) <= 1 for i,v in enumerate(A)) - + diff --git a/Python/gray-code.py b/Python/gray-code.py index 415e0f752..f54b9b421 100644 --- a/Python/gray-code.py +++ b/Python/gray-code.py @@ -2,22 +2,22 @@ # Space: O(1) # The gray code is a binary numeral system where two successive values differ in only one bit. -# -# Given a non-negative integer n representing the total number of bits in the code, +# +# Given a non-negative integer n representing the total number of bits in the code, # print the sequence of gray code. A gray code sequence must begin with 0. -# +# # For example, given n = 2, return [0,1,3,2]. Its gray code sequence is: -# +# # 00 - 0 # 01 - 1 # 11 - 3 # 10 - 2 # Note: # For a given n, a gray code sequence is not uniquely defined. -# +# # For example, [0,2,3,1] is also a valid gray code sequence according # to the above definition. -# +# # For now, the judge is able to judge based on one instance of gray code # sequence. Sorry about that. diff --git a/Python/guess-number-higher-or-lower-ii.py b/Python/guess-number-higher-or-lower-ii.py index 48e7139a8..9de4bf3bc 100644 --- a/Python/guess-number-higher-or-lower-ii.py +++ b/Python/guess-number-higher-or-lower-ii.py @@ -26,7 +26,7 @@ # Hint: # # The best strategy to play the game is to minimize the maximum loss -# you could possibly face. Another strategy is to minimize the expected loss. +# you could possibly face. Another strategy is to minimize the expected loss. # Here, we are interested in the first scenario. # Take a small example (n = 3). What do you end up paying in the worst case? # Check out this article if you're still stuck. diff --git a/Python/h-index.py b/Python/h-index.py index 4ad07afdb..ab47ad881 100644 --- a/Python/h-index.py +++ b/Python/h-index.py @@ -4,15 +4,15 @@ # Given an array of citations (each citation is a non-negative integer) # of a researcher, write a function to compute the researcher's h-index. # -# According to the definition of h-index on Wikipedia: +# According to the definition of h-index on Wikipedia: # "A scientist has index h if h of his/her N papers have # at least h citations each, and the other N − h papers have # no more than h citations each." # -# For example, given citations = [3, 0, 6, 1, 5], +# For example, given citations = [3, 0, 6, 1, 5], # which means the researcher has 5 papers in total -# and each of them had received 3, 0, 6, 1, 5 citations respectively. -# Since the researcher has 3 papers with at least 3 citations each and +# and each of them had received 3, 0, 6, 1, 5 citations respectively. +# Since the researcher has 3 papers with at least 3 citations each and # the remaining two with no more than 3 citations each, his h-index is 3. # # Note: If there are several possible values for h, the maximum one is taken as the h-index. @@ -57,7 +57,7 @@ def hIndex(self, citations): else: break return h - + # Time: O(nlogn) # Space: O(n) class Solution3(object): @@ -67,4 +67,4 @@ def hIndex(self, citations): :rtype: int """ return sum(x >= i + 1 for i, x in enumerate(sorted(citations, reverse=True))) - + diff --git a/Python/happy-number.py b/Python/happy-number.py index 7ea4a5675..588f655e5 100644 --- a/Python/happy-number.py +++ b/Python/happy-number.py @@ -3,11 +3,11 @@ # # Write an algorithm to determine if a number is "happy". # -# A happy number is a number defined by the following process: -# Starting with any positive integer, replace the number by the sum -# of the squares of its digits, and repeat the process until -# the number equals 1 (where it will stay), or it loops endlessly -# in a cycle which does not include 1. Those numbers for which +# A happy number is a number defined by the following process: +# Starting with any positive integer, replace the number by the sum +# of the squares of its digits, and repeat the process until +# the number equals 1 (where it will stay), or it loops endlessly +# in a cycle which does not include 1. Those numbers for which # this process ends in 1 are happy numbers. # # Example: 19 is a happy number @@ -26,7 +26,7 @@ def isHappy(self, n): lookup[n] = True n = self.nextNumber(n) return n == 1 - + def nextNumber(self, n): new = 0 for char in str(n): diff --git a/Python/house-robber-ii.py b/Python/house-robber-ii.py index 554ed4828..5f1b2fbff 100644 --- a/Python/house-robber-ii.py +++ b/Python/house-robber-ii.py @@ -3,13 +3,13 @@ # # Note: This is an extension of House Robber. # -# After robbing those houses on that street, the thief has found himself a new place -# for his thievery so that he will not get too much attention. This time, all houses -# at this place are arranged in a circle. That means the first house is the neighbor -# of the last one. Meanwhile, the security system for these houses remain the same as +# After robbing those houses on that street, the thief has found himself a new place +# for his thievery so that he will not get too much attention. This time, all houses +# at this place are arranged in a circle. That means the first house is the neighbor +# of the last one. Meanwhile, the security system for these houses remain the same as # for those in the previous street. # -# Given a list of non-negative integers representing the amount of money of each house, +# Given a list of non-negative integers representing the amount of money of each house, # determine the maximum amount of money you can rob tonight without alerting the police. # class Solution: @@ -18,19 +18,19 @@ class Solution: def rob(self, nums): if len(nums) == 0: return 0 - + if len(nums) == 1: return nums[0] - + return max(self.robRange(nums, 0, len(nums) - 1),\ self.robRange(nums, 1, len(nums))) - + def robRange(self, nums, start, end): num_i, num_i_1 = nums[start], 0 for i in xrange(start + 1, end): num_i_1, num_i_2 = num_i, num_i_1 num_i = max(nums[i] + num_i_2, num_i_1); - + return num_i if __name__ == '__main__': diff --git a/Python/house-robber-iii.py b/Python/house-robber-iii.py index 06d4ea317..a1ed8c3a1 100644 --- a/Python/house-robber-iii.py +++ b/Python/house-robber-iii.py @@ -2,11 +2,11 @@ # Space: O(h) # The thief has found himself a new place for his thievery again. -# There is only one entrance to this area, called the "root." +# There is only one entrance to this area, called the "root." # Besides the root, each house has one and only one parent house. # After a tour, the smart thief realized that "all houses in this # place forms a binary tree". It will automatically contact the -# police if two directly-linked houses were broken into on the +# police if two directly-linked houses were broken into on the # same night. # # Determine the maximum amount of money the thief can rob tonight @@ -16,14 +16,14 @@ # 3 # / \ # 2 3 -# \ \ +# \ \ # 3 1 # Maximum amount of money the thief can rob = 3 + 3 + 1 = 7. # Example 2: # 3 # / \ # 4 5 -# / \ \ +# / \ \ # 1 3 1 # Maximum amount of money the thief can rob = 4 + 5 = 9. # @@ -45,5 +45,5 @@ def robHelper(root): return (0, 0) left, right = robHelper(root.left), robHelper(root.right) return (root.val + left[1] + right[1], max(left) + max(right)) - + return max(robHelper(root)) diff --git a/Python/house-robber.py b/Python/house-robber.py index 3efd6e8a0..d94ae3b53 100644 --- a/Python/house-robber.py +++ b/Python/house-robber.py @@ -1,12 +1,12 @@ # Time: O(n) # Space: O(1) # -# You are a professional robber planning to rob houses along a street. -# Each house has a certain amount of money stashed, the only constraint stopping you +# You are a professional robber planning to rob houses along a street. +# Each house has a certain amount of money stashed, the only constraint stopping you # from robbing each of them is that adjacent houses have security system connected # and it will automatically contact the police if two adjacent houses were broken into on the same night. # -# Given a list of non-negative integers representing the amount of money of each house, +# Given a list of non-negative integers representing the amount of money of each house, # determine the maximum amount of money you can rob tonight without alerting the police. # class Solution: @@ -15,15 +15,15 @@ class Solution: def rob(self, num): if len(num) == 0: return 0 - + if len(num) == 1: return num[0] - + num_i, num_i_1 = max(num[1], num[0]), num[0] for i in xrange(2, len(num)): num_i_1, num_i_2 = num_i, num_i_1 num_i = max(num[i] + num_i_2, num_i_1); - + return num_i def rob2(self, nums): diff --git a/Python/image-smoother.py b/Python/image-smoother.py index 232662103..cd51b2fbc 100644 --- a/Python/image-smoother.py +++ b/Python/image-smoother.py @@ -22,7 +22,7 @@ # Note: # The value in the given matrix is in the range of [0, 255]. # The length and width of the given matrix are in the range of [1, 150]. - + class Solution(object): def imageSmoother(self, M): """ diff --git a/Python/implement-queue-using-stacks.py b/Python/implement-queue-using-stacks.py index 02b3e4860..9c32c8197 100644 --- a/Python/implement-queue-using-stacks.py +++ b/Python/implement-queue-using-stacks.py @@ -32,14 +32,14 @@ def push(self, x): def pop(self): self.peek() return self.B.pop() - + # @return an integer def peek(self): if not self.B: while self.A: self.B.append(self.A.pop()) return self.B[-1] - + # @return an boolean def empty(self): return not self.A and not self.B diff --git a/Python/implement-strstr.py b/Python/implement-strstr.py index a592a3654..1e7a5919a 100644 --- a/Python/implement-strstr.py +++ b/Python/implement-strstr.py @@ -2,7 +2,7 @@ # Space: O(k) # # Implement strStr(). -# +# # Returns a pointer to the first occurrence of needle in haystack, # or null if needle is not part of haystack. # @@ -18,9 +18,9 @@ def strStr(self, haystack, needle): """ if not needle: return 0 - + return self.KMP(haystack, needle) - + def KMP(self, text, pattern): prefix = self.getPrefix(pattern) j = -1 @@ -32,7 +32,7 @@ def KMP(self, text, pattern): if j == len(pattern) - 1: return i - j return -1 - + def getPrefix(self, pattern): prefix = [-1] * len(pattern) j = -1 @@ -44,7 +44,7 @@ def getPrefix(self, pattern): prefix[i] = j return prefix - + # Time: O(n * k) # Space: O(k) class Solution2(object): diff --git a/Python/implement-trie-prefix-tree.py b/Python/implement-trie-prefix-tree.py index 0301555ad..a062f91ba 100644 --- a/Python/implement-trie-prefix-tree.py +++ b/Python/implement-trie-prefix-tree.py @@ -12,7 +12,7 @@ class TrieNode: def __init__(self): self.is_string = False self.leaves = {} - + class Trie: @@ -37,7 +37,7 @@ def search(self, word): node = self.childSearch(word) if node: return node.is_string - return False + return False # @param {string} prefix # @return {boolean} diff --git a/Python/increasing-subsequences.py b/Python/increasing-subsequences.py index e246be299..6639959bc 100644 --- a/Python/increasing-subsequences.py +++ b/Python/increasing-subsequences.py @@ -1,8 +1,8 @@ # Time: O(n * 2^n) # Space: O(n^2) -# Given an integer array, your task is -# to find all the different possible increasing +# Given an integer array, your task is +# to find all the different possible increasing # subsequences of the given array, # and the length of an increasing subsequence should be at least 2 . # diff --git a/Python/insert-delete-getrandom-o1-duplicates-allowed.py b/Python/insert-delete-getrandom-o1-duplicates-allowed.py index 3aa6ab1a7..f0dbf9c63 100644 --- a/Python/insert-delete-getrandom-o1-duplicates-allowed.py +++ b/Python/insert-delete-getrandom-o1-duplicates-allowed.py @@ -43,7 +43,7 @@ def __init__(self): """ self.__list = [] self.__used = defaultdict(list) - + def insert(self, val): """ diff --git a/Python/insert-delete-getrandom-o1.py b/Python/insert-delete-getrandom-o1.py index 136636091..9ae83a0be 100644 --- a/Python/insert-delete-getrandom-o1.py +++ b/Python/insert-delete-getrandom-o1.py @@ -45,7 +45,7 @@ def __init__(self): """ self.__set = [] self.__used = {} - + def insert(self, val): """ @@ -60,7 +60,7 @@ def insert(self, val): self.__used[val] = len(self.__set)-1 return True - + def remove(self, val): """ diff --git a/Python/insert-interval.py b/Python/insert-interval.py index 16cf7c0a7..33dd83cec 100644 --- a/Python/insert-interval.py +++ b/Python/insert-interval.py @@ -4,13 +4,13 @@ # Given a set of non-overlapping intervals, insert a new interval into the # intervals (merge if necessary). # You may assume that the intervals were initially sorted according to their start times. -# +# # Example 1: # Given intervals [1,3],[6,9], insert and merge [2,5] in as [1,5],[6,9]. -# +# # Example 2: # Given [1,2],[3,5],[6,7],[8,10],[12,16], insert and merge [4,9] in as [1,2],[3,10],[12,16]. -# +# # This is because the new interval [4,9] overlaps with [3,5],[6,7],[8,10]. # Definition for an interval. @@ -18,7 +18,7 @@ class Interval: def __init__(self, s=0, e=0): self.start = s self.end = e - + def __repr__(self): return "[{}, {}]".format(self.start, self.end) diff --git a/Python/insertion-sort-list.py b/Python/insertion-sort-list.py index fa7c9616c..30a8a7500 100644 --- a/Python/insertion-sort-list.py +++ b/Python/insertion-sort-list.py @@ -9,7 +9,7 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) @@ -22,7 +22,7 @@ class Solution: def insertionSortList(self, head): if head is None or self.isSorted(head): return head - + dummy = ListNode(-2147483648) dummy.next = head cur, sorted_tail = head.next, head @@ -31,13 +31,13 @@ def insertionSortList(self, head): while prev.next.val < cur.val: prev = prev.next if prev == sorted_tail: - cur, sorted_tail = cur.next, cur + cur, sorted_tail = cur.next, cur else: cur.next, prev.next, sorted_tail.next = prev.next, cur, cur.next cur = sorted_tail.next - + return dummy.next - + def isSorted(self, head): while head and head.next: if head.val > head.next.val: diff --git a/Python/integer-break.py b/Python/integer-break.py index c2127f3e9..ca3bab0e5 100644 --- a/Python/integer-break.py +++ b/Python/integer-break.py @@ -31,15 +31,15 @@ def integerBreak(self, n): # ai <= 2 * (ai - 2) # - For each aj >= 5, we can always maximize the product by: # aj <= 3 * (aj - 3) - # + # # Conclusion 1: # - For n >= 4, the max of the product must be in the form of # 3^a * 2^b, s.t. 3a + 2b = n - # + # # 2. To maximize the product = 3^a * 2^b s.t. 3a + 2b = n # - For each b >= 3, we can always maximize the product by: # 3^a * 2^b <= 3^(a+2) * 2^(b-3) s.t. 3(a+2) + 2(b-3) = n - # + # # Conclusion 2: # - For n >= 4, the max of the product must be in the form of # 3^Q * 2^R, 0 <= R < 3 s.t. 3Q + 2R = n diff --git a/Python/integer-replacement.py b/Python/integer-replacement.py index 8ec74bf04..4820bf85a 100644 --- a/Python/integer-replacement.py +++ b/Python/integer-replacement.py @@ -49,9 +49,9 @@ def integerReplacement(self, n): else: n /= 2 result += 1 - + return result - + # Time: O(logn) # Space: O(logn) @@ -70,4 +70,4 @@ def integerReplacement(self, n): return self.integerReplacement((n - 1) / 4) + 3 else: return self.integerReplacement((n + 1) / 4) + 3 - + diff --git a/Python/integer-to-english-words.py b/Python/integer-to-english-words.py index 3c5ab64db..6097cf02f 100644 --- a/Python/integer-to-english-words.py +++ b/Python/integer-to-english-words.py @@ -57,7 +57,7 @@ def threeDigits(self, num, lookup, unit): if unit != "": res.append(unit) return " ".join(res) - + def twoDigits(self, num, lookup): if num in lookup: return lookup[num] diff --git a/Python/integer-to-roman.py b/Python/integer-to-roman.py index eed075b1b..672201a80 100644 --- a/Python/integer-to-roman.py +++ b/Python/integer-to-roman.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given an integer, convert it to a roman numeral. -# +# # Input is guaranteed to be within the range from 1 to 3999. # @@ -17,16 +17,16 @@ def intToRoman(self, num): 100: "C", 400: "CD", 500: "D", 900: "CM", \ 1000: "M"} keyset, result = sorted(numeral_map.keys()), [] - + while num > 0: for key in reversed(keyset): while num / key > 0: num -= key result += numeral_map[key] - + return "".join(result) - + if __name__ == "__main__": print Solution().intToRoman(999) print Solution().intToRoman(3999) diff --git a/Python/interleaving-string.py b/Python/interleaving-string.py index fd92f6141..8e0ff98c6 100644 --- a/Python/interleaving-string.py +++ b/Python/interleaving-string.py @@ -2,16 +2,16 @@ # Space: O(m + n) # # Given s1, s2, s3, find whether s3 is formed by the interleaving of s1 and s2. -# +# # For example, # Given: # s1 = "aabcc", # s2 = "dbbca", -# +# # When s3 = "aadbbcbcac", return true. # When s3 = "aadbbbaccc", return false. # - + # Time: O(m * n) # Space: O(m + n) # Dynamic Programming + Sliding Window @@ -55,7 +55,7 @@ def isInterleave(self, s1, s2, s3): # Time: O(m * n) # Space: O(m * n) -# Recursive + Hash +# Recursive + Hash class Solution3: # @return a boolean def isInterleave(self, s1, s2, s3): @@ -63,22 +63,22 @@ def isInterleave(self, s1, s2, s3): if len(s1) + len(s2) != len(s3): return False return self.isInterleaveRecu(s1, s2, s3, 0, 0, 0) - + def isInterleaveRecu(self, s1, s2, s3, a, b, c): if repr([a, b]) in self.match.keys(): return self.match[repr([a, b])] - + if c == len(s3): return True - + result = False if a < len(s1) and s1[a] == s3[c]: result = result or self.isInterleaveRecu(s1, s2, s3, a + 1, b, c + 1) if b < len(s2) and s2[b] == s3[c]: result = result or self.isInterleaveRecu(s1, s2, s3, a, b + 1, c + 1) - - self.match[repr([a, b])] = result - + + self.match[repr([a, b])] = result + return result if __name__ == "__main__": diff --git a/Python/intersection-of-two-arrays-ii.py b/Python/intersection-of-two-arrays-ii.py index 9e3f626c6..df7776dca 100644 --- a/Python/intersection-of-two-arrays-ii.py +++ b/Python/intersection-of-two-arrays-ii.py @@ -44,7 +44,7 @@ def intersect(self, nums1, nums2): """ if len(nums1) > len(nums2): return self.intersect(nums2, nums1) - + lookup = collections.defaultdict(int) for i in nums1: lookup[i] += 1 @@ -102,7 +102,7 @@ def binary_search(compare, nums, left, right, target): if left != len(nums2) and nums2[left] == i: res += i, left += 1 - + return res @@ -120,7 +120,7 @@ def intersect(self, nums1, nums2): nums1.sort(), nums2.sort() # Make sure it is sorted, doesn't count in time. res = [] - + it1, it2 = 0, 0 while it1 < len(nums1) and it2 < len(nums2): if nums1[it1] < nums2[it2]: @@ -131,7 +131,7 @@ def intersect(self, nums1, nums2): res += nums1[it1], it1 += 1 it2 += 1 - + return res @@ -149,7 +149,7 @@ def intersect(self, nums1, nums2): nums1.sort(), nums2.sort() # O(max(m, n) * log(max(m, n))) res = [] - + it1, it2 = 0, 0 while it1 < len(nums1) and it2 < len(nums2): if nums1[it1] < nums2[it2]: @@ -160,5 +160,5 @@ def intersect(self, nums1, nums2): res += nums1[it1], it1 += 1 it2 += 1 - + return res diff --git a/Python/intersection-of-two-arrays.py b/Python/intersection-of-two-arrays.py index e04ff0c40..fb90f68eb 100644 --- a/Python/intersection-of-two-arrays.py +++ b/Python/intersection-of-two-arrays.py @@ -22,7 +22,7 @@ def intersection(self, nums1, nums2): """ if len(nums1) > len(nums2): return self.intersection(nums2, nums1) - + lookup = set() for i in nums1: lookup.add(i) @@ -75,7 +75,7 @@ def binary_search(compare, nums, left, right, target): if left != len(nums2) and nums2[left] == i: res += i, left = binary_search(lambda x, y: x > y, nums2, left, len(nums2), i) - + return res @@ -91,7 +91,7 @@ def intersection(self, nums1, nums2): """ nums1.sort(), nums2.sort() res = [] - + it1, it2 = 0, 0 while it1 < len(nums1) and it2 < len(nums2): if nums1[it1] < nums2[it2]: @@ -103,5 +103,5 @@ def intersection(self, nums1, nums2): res += nums1[it1], it1 += 1 it2 += 1 - + return res diff --git a/Python/intersection-of-two-linked-lists.py b/Python/intersection-of-two-linked-lists.py index 328df54fb..3d18b7918 100644 --- a/Python/intersection-of-two-linked-lists.py +++ b/Python/intersection-of-two-linked-lists.py @@ -2,20 +2,20 @@ # Space: O(1) # # Write a program to find the node at which the intersection of two singly linked lists begins. -# -# +# +# # For example, the following two linked lists: -# +# # A: a1 - a2 # \ # c1 - c2 - c3 -# / +# / # B: b1 - b2 - b3 # begin to intersect at node c1. -# -# +# +# # Notes: -# +# # If the two linked lists have no intersection at all, return null. # The linked lists must retain their original structure after the function returns. # You may assume there are no cycles anywhere in the entire linked structure. @@ -34,14 +34,14 @@ class Solution: def getIntersectionNode(self, headA, headB): curA, curB = headA, headB begin, tailA, tailB = None, None, None - + # a->c->b->c # b->c->a->c while curA and curB: if curA == curB: begin = curA break - + if curA.next: curA = curA.next elif tailA is None: @@ -49,7 +49,7 @@ def getIntersectionNode(self, headA, headB): curA = headB else: break - + if curB.next: curB = curB.next elif tailB is None: @@ -57,5 +57,5 @@ def getIntersectionNode(self, headA, headB): curB = headA else: break - + return begin diff --git a/Python/ip-to-cidr.py b/Python/ip-to-cidr.py index 3c4bb3b6f..fe9a17788 100644 --- a/Python/ip-to-cidr.py +++ b/Python/ip-to-cidr.py @@ -39,7 +39,7 @@ # but our answer was the shortest possible. # # Also note that a representation beginning with say, "255.0.0.7/30" would be incorrect, -# because it includes addresses like 255.0.0.4 = 11111111 00000000 00000000 00000100 +# because it includes addresses like 255.0.0.4 = 11111111 00000000 00000000 00000100 # that are outside the specified range. # Note: # - ip will be a valid IPv4 address. diff --git a/Python/is-graph-bipartite.py b/Python/is-graph-bipartite.py index 213d8e372..5b9890e91 100644 --- a/Python/is-graph-bipartite.py +++ b/Python/is-graph-bipartite.py @@ -8,7 +8,7 @@ # one node in A and another node in B. # # The graph is given in the following form: graph[i] is a list of indexes j -# for which the edge between nodes i and j exists. +# for which the edge between nodes i and j exists. # Each node is an integer between 0 and graph.length - 1. # There are no self edges or parallel edges: graph[i] does not contain i, # and it doesn't contain any element twice. @@ -16,18 +16,18 @@ # Example 1: # Input: [[1,3], [0,2], [1,3], [0,2]] # Output: true -# Explanation: +# Explanation: # The graph looks like this: # 0----1 # | | # | | # 3----2 # We can divide the vertices into two groups: {0, 2} and {1, 3}. -# +# # Example 2: # Input: [[1,2,3], [0,2], [0,1,3], [0,2]] # Output: false -# Explanation: +# Explanation: # The graph looks like this: # 0----1 # | \ | diff --git a/Python/is-subsequence.py b/Python/is-subsequence.py index e92eceef0..a47504b7d 100644 --- a/Python/is-subsequence.py +++ b/Python/is-subsequence.py @@ -6,7 +6,7 @@ # You may assume that there is only lower case English letters in both s and t. # t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100). # -# A subsequence of a string is a new string which is formed from +# A subsequence of a string is a new string which is formed from # the original string by deleting some (can be none) of the characters # without disturbing the relative positions of the remaining characters. # (ie, "ace" is a subsequence of "abcde" while "aec" is not). diff --git a/Python/isomorphic-strings.py b/Python/isomorphic-strings.py index e22243e67..4268ab641 100644 --- a/Python/isomorphic-strings.py +++ b/Python/isomorphic-strings.py @@ -5,8 +5,8 @@ # # Two strings are isomorphic if the characters in s can be replaced to get t. # -# All occurrences of a character must be replaced with another character -# while preserving the order of characters. No two characters may map to +# All occurrences of a character must be replaced with another character +# while preserving the order of characters. No two characters may map to # the same character but a character may map to itself. # # For example, @@ -35,7 +35,7 @@ def isIsomorphic(self, s, t): for p, w in izip(s, t): if w not in s2t and p not in t2s: s2t[w] = p - t2s[p] = w + t2s[p] = w elif w not in s2t or s2t[w] != p: # Contradict mapping. return False @@ -48,7 +48,7 @@ class Solution2(object): def isIsomorphic(self, s, t): if len(s) != len(t): return False - + return self.halfIsom(s, t) and self.halfIsom(t, s) def halfIsom(self, s, t): diff --git a/Python/jewels-and-stones.py b/Python/jewels-and-stones.py index 18fbbd812..61fa6936d 100644 --- a/Python/jewels-and-stones.py +++ b/Python/jewels-and-stones.py @@ -30,4 +30,4 @@ def numJewelsInStones(self, J, S): """ lookup = set(J) return sum(s in lookup for s in S) - + diff --git a/Python/judge-route-circle.py b/Python/judge-route-circle.py index 79cfe6d5d..8aafd1fee 100644 --- a/Python/judge-route-circle.py +++ b/Python/judge-route-circle.py @@ -14,7 +14,7 @@ # Example 2: # Input: "LL" # Output: false - + class Solution(object): def judgeCircle(self, moves): """ diff --git a/Python/jump-game-ii.py b/Python/jump-game-ii.py index 571c405d4..d99f97642 100644 --- a/Python/jump-game-ii.py +++ b/Python/jump-game-ii.py @@ -2,14 +2,14 @@ # Space: O(1) # # Given an array of non-negative integers, you are initially positioned at the first index of the array. -# +# # Each element in the array represents your maximum jump length at that position. -# +# # Your goal is to reach the last index in the minimum number of jumps. -# +# # For example: # Given array A = [2,3,1,1,4] -# +# # The minimum number of jumps to reach the last index is 2. (Jump 1 step from index 0 to 1, then 3 steps to the last index.) # @@ -29,8 +29,8 @@ def jump(self, A): jump_count += 1 reachable = max(reachable, i + length) return jump_count - - + + if __name__ == "__main__": print Solution().jump([2,3,1,1,4]) print Solution().jump([3,2,1,0,4]) diff --git a/Python/jump-game.py b/Python/jump-game.py index f3e18cce6..de25a28b3 100644 --- a/Python/jump-game.py +++ b/Python/jump-game.py @@ -2,14 +2,14 @@ # Space: O(1) # # Given an array of non-negative integers, you are initially positioned at the first index of the array. -# +# # Each element in the array represents your maximum jump length at that position. -# +# # Determine if you are able to reach the last index. -# +# # For example: # A = [2,3,1,1,4], return true. -# +# # A = [3,2,1,0,4], return false. class Solution: @@ -22,7 +22,7 @@ def canJump(self, A): break reachable = max(reachable, i + length) return reachable >= len(A) - 1 - + if __name__ == "__main__": print Solution().canJump([2,3,1,1,4]) print Solution().canJump([3,2,1,0,4]) \ No newline at end of file diff --git a/Python/k-diff-pairs-in-an-array.py b/Python/k-diff-pairs-in-an-array.py index 58a4b29b5..9331f8a74 100644 --- a/Python/k-diff-pairs-in-an-array.py +++ b/Python/k-diff-pairs-in-an-array.py @@ -5,7 +5,7 @@ # Total Submissions: 20941 # Difficulty: Easy # Contributors: murali.kf370 -# Given an array of integers and an integer k, +# Given an array of integers and an integer k, # you need to find the number of unique k-diff pairs in the array. # Here a k-diff pair is defined as an integer pair (i, j), # where i and j are both numbers in the array and their absolute difference is k. diff --git a/Python/k-empty-slots.py b/Python/k-empty-slots.py index ca99f58b1..c87e9005a 100644 --- a/Python/k-empty-slots.py +++ b/Python/k-empty-slots.py @@ -18,14 +18,14 @@ # If there isn't such day, output -1. # # Example 1: -# Input: +# Input: # flowers: [1,3,2] # k: 1 # Output: 2 # Explanation: In the second day, the first and the third flower have become blooming. # # Example 2: -# Input: +# Input: # flowers: [1,2,3] # k: 1 # Output: -1 @@ -45,10 +45,10 @@ def kEmptySlots(self, flowers, k): result = float("inf") i, left, right = 0, 0, k+1 while right < len(days): - if days[i] < days[left] or days[i] <= days[right]: + if days[i] < days[left] or days[i] <= days[right]: if i == right: result = min(result, max(days[left], days[right])) left, right = i, k+1+i; i += 1 return -1 if result == float("inf") else result+1 - + diff --git a/Python/k-inverse-pairs-array.py b/Python/k-inverse-pairs-array.py index da1e592dd..1845a3c44 100644 --- a/Python/k-inverse-pairs-array.py +++ b/Python/k-inverse-pairs-array.py @@ -12,12 +12,12 @@ # Example 1: # Input: n = 3, k = 0 # Output: 1 -# Explanation: +# Explanation: # Only the array [1,2,3] which consists of numbers from 1 to 3 has exactly 0 inverse pair. # Example 2: # Input: n = 3, k = 1 # Output: 2 -# Explanation: +# Explanation: # The array [1,3,2] and [2,1,3] have exactly 1 inverse pair. # Note: # The integer n is in the range [1, 1000] and k is in the range [0, 1000]. diff --git a/Python/k-th-smallest-in-lexicographical-order.py b/Python/k-th-smallest-in-lexicographical-order.py index 4a969c6d5..8a15760be 100644 --- a/Python/k-th-smallest-in-lexicographical-order.py +++ b/Python/k-th-smallest-in-lexicographical-order.py @@ -78,7 +78,7 @@ def count(n, prefix): number *= 10 result -= max(number/10 - (n - prefix/10 + 1), 0) return result - + def findKthNumberHelper(n, k, cur, index): if cur: index += 1 @@ -98,5 +98,5 @@ def findKthNumberHelper(n, k, cur, index): i += 1 cur /= 10 return (0, index) - + return findKthNumberHelper(n, k, 0, 0)[0] diff --git a/Python/k-th-symbol-in-grammar.py b/Python/k-th-symbol-in-grammar.py index dbed8a5f6..8980720a1 100644 --- a/Python/k-th-symbol-in-grammar.py +++ b/Python/k-th-symbol-in-grammar.py @@ -45,5 +45,5 @@ def bitCount(n): n &= n - 1 result += 1 return result - + return bitCount(K-1) % 2 diff --git a/Python/keyboard-row.py b/Python/keyboard-row.py index 9291dcf63..ef825d00c 100644 --- a/Python/keyboard-row.py +++ b/Python/keyboard-row.py @@ -18,7 +18,7 @@ def findWords(self, words): :rtype: List[str] """ rows = [set(['q', 'w', 'e', 'r', 't', 'y','u', 'i', 'o', 'p']), - set(['a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l']), + set(['a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l']), set(['z', 'x', 'c', 'v', 'b' ,'n', 'm'])] result = [] diff --git a/Python/knight-probability-in-chessboard.py b/Python/knight-probability-in-chessboard.py index cb87d07ed..2f493ca96 100644 --- a/Python/knight-probability-in-chessboard.py +++ b/Python/knight-probability-in-chessboard.py @@ -1,6 +1,6 @@ # Time: O(k * n^2) # Space: O(n^2) - + # On an NxN chessboard, a knight starts at the r-th row and c-th column and # attempts to make exactly K moves. The rows and columns are 0 indexed, # so the top-left square is (0, 0), and the bottom-right square is (N-1, N-1). @@ -17,7 +17,7 @@ # Example: # Input: 3, 2, 0, 0 # Output: 0.0625 -# +# # Explanation: There are two moves (to (1,2), (2,1)) that will keep the knight on the board. # From each of those positions, there are also two moves that will keep the knight on the board. # The total probability the knight stays on the board is 0.0625. diff --git a/Python/kth-largest-element-in-an-array.py b/Python/kth-largest-element-in-an-array.py index 08dd0b788..55e1af2a9 100644 --- a/Python/kth-largest-element-in-an-array.py +++ b/Python/kth-largest-element-in-an-array.py @@ -18,7 +18,7 @@ def findKthLargest(self, nums, k): right = new_pivot_idx - 1 else: # new_pivot_idx < k - 1. left = new_pivot_idx + 1 - + def PartitionAroundPivot(self, left, right, pivot_idx, nums): pivot_value = nums[pivot_idx] new_pivot_idx = left @@ -27,7 +27,7 @@ def PartitionAroundPivot(self, left, right, pivot_idx, nums): if nums[i] > pivot_value: nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i] new_pivot_idx += 1 - + nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right] return new_pivot_idx - + diff --git a/Python/kth-smallest-element-in-a-bst.py b/Python/kth-smallest-element-in-a-bst.py index ee7216094..89457f4e9 100644 --- a/Python/kth-smallest-element-in-a-bst.py +++ b/Python/kth-smallest-element-in-a-bst.py @@ -3,7 +3,7 @@ # Given a binary search tree, write a function kthSmallest to find the kth smallest element in it. # -# Note: +# Note: # You may assume k is always valid, 1 ≤ k ≤ BST's total elements. # # Follow up: diff --git a/Python/kth-smallest-element-in-a-sorted-matrix.py b/Python/kth-smallest-element-in-a-sorted-matrix.py index be66a8b8f..1bf17df12 100644 --- a/Python/kth-smallest-element-in-a-sorted-matrix.py +++ b/Python/kth-smallest-element-in-a-sorted-matrix.py @@ -18,7 +18,7 @@ # k = 8, # # return 13. -# Note: +# Note: # You may assume k is always valid, 1 <= k <= n^2. from heapq import heappush, heappop diff --git a/Python/kth-smallest-number-in-multiplication-table.py b/Python/kth-smallest-number-in-multiplication-table.py index dc5ee176e..17c4225b8 100644 --- a/Python/kth-smallest-number-in-multiplication-table.py +++ b/Python/kth-smallest-number-in-multiplication-table.py @@ -9,8 +9,8 @@ # # Example 1: # Input: m = 3, n = 3, k = 5 -# Output: -# Explanation: +# Output: +# Explanation: # The Multiplication Table: # 1 2 3 # 2 4 6 @@ -19,8 +19,8 @@ # The 5-th smallest number is 3 (1, 2, 2, 3, 3). # Example 2: # Input: m = 2, n = 3, k = 6 -# Output: -# Explanation: +# Output: +# Explanation: # The Multiplication Table: # 1 2 3 # 2 4 6 diff --git a/Python/largest-bst-subtree.py b/Python/largest-bst-subtree.py index 35d18f1be..78359e2bb 100644 --- a/Python/largest-bst-subtree.py +++ b/Python/largest-bst-subtree.py @@ -25,7 +25,7 @@ def largestBSTSubtreeHelper(root): left_size, left_min, left_max = 0, root.val, root.val if root.left is not None: left_size, left_min, left_max = largestBSTSubtreeHelper(root.left) - + right_size, right_min, right_max = 0, root.val, root.val if root.right is not None: right_size, right_min, right_max = largestBSTSubtreeHelper(root.right) diff --git a/Python/largest-number.py b/Python/largest-number.py index 8317a617b..008d5e0e7 100644 --- a/Python/largest-number.py +++ b/Python/largest-number.py @@ -2,9 +2,9 @@ # Space: O(1) # # Given a list of non negative integers, arrange them such that they form the largest number. -# +# # For example, given [3, 30, 34, 5, 9], the largest formed number is 9534330. -# +# # Note: The result may be very large, so you need to return a string instead of an integer. # @@ -19,4 +19,4 @@ def largestNumber(self, num): if __name__ == "__main__": num = [3, 30, 34, 5, 9] - print Solution().largestNumber(num) + print Solution().largestNumber(num) diff --git a/Python/largest-palindrome-product.py b/Python/largest-palindrome-product.py index c189a0011..66c4b3fee 100644 --- a/Python/largest-palindrome-product.py +++ b/Python/largest-palindrome-product.py @@ -20,7 +20,7 @@ def largestPalindrome(self, n): """ if n == 1: return 9 - + upper, lower = 10**n-1, 10**(n-1) for i in reversed(xrange(lower, upper+1)): candidate = int(str(i) + str(i)[::-1]) diff --git a/Python/largest-plus-sign.py b/Python/largest-plus-sign.py index 5b47e514b..3d068753c 100644 --- a/Python/largest-plus-sign.py +++ b/Python/largest-plus-sign.py @@ -20,7 +20,7 @@ def orderOfLargestPlusSign(self, N, mines): for j in reversed(xrange(N)): l = 0 if (i, j) in lookup else l+1 dp[i][j] = min(dp[i][j], l) - + for j in xrange(N): l = 0 for i in xrange(N): diff --git a/Python/largest-rectangle-in-histogram.py b/Python/largest-rectangle-in-histogram.py index e3dea568b..3e4d6fa82 100644 --- a/Python/largest-rectangle-in-histogram.py +++ b/Python/largest-rectangle-in-histogram.py @@ -1,10 +1,10 @@ # Time: O(n) # Space: O(n) # -# Given n non-negative integers representing the histogram's bar +# Given n non-negative integers representing the histogram's bar # height where the width of each bar is 1, # find the area of largest rectangle in the histogram. -# +# # For example, # Given height = [2,1,5,6,2,3], # return 10. @@ -30,4 +30,4 @@ def largestRectangleArea(self, height): if __name__ == "__main__": print Solution().largestRectangleArea([2, 0, 2]) print Solution().largestRectangleArea([2, 1, 5, 6, 2, 3]) - + diff --git a/Python/length-of-last-word.py b/Python/length-of-last-word.py index 0e713d1a1..ec1c239e0 100644 --- a/Python/length-of-last-word.py +++ b/Python/length-of-last-word.py @@ -2,12 +2,12 @@ # Space: O(1) # # Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string. -# +# # If the last word does not exist, return 0. -# +# # Note: A word is defined as a character sequence consists of non-space characters only. -# -# For example, +# +# For example, # Given s = "Hello World", # return 5. # diff --git a/Python/letter-combinations-of-a-phone-number.py b/Python/letter-combinations-of-a-phone-number.py index c876137ca..ca3d1c347 100644 --- a/Python/letter-combinations-of-a-phone-number.py +++ b/Python/letter-combinations-of-a-phone-number.py @@ -2,11 +2,11 @@ # Space: O(n) # # Given a digit string, return all possible letter combinations that the number could represent. -# +# # A mapping of digit to letters (just like on the telephone buttons) is given below. -# +# # lookup = ["", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"] -# +# # Input:Digit string "23" # Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. # Note: @@ -19,18 +19,18 @@ class Solution: def letterCombinations(self, digits): if not digits: return [] - + lookup, result = ["", "", "abc", "def", "ghi", "jkl", "mno", \ "pqrs", "tuv", "wxyz"], [""] for digit in reversed(digits): choices = lookup[int(digit)] m, n = len(choices), len(result) - result += [result[i % n] for i in xrange(n, m * n)] + result += [result[i % n] for i in xrange(n, m * n)] for i in xrange(m * n): - result[i] = choices[i / n] + result[i] - + result[i] = choices[i / n] + result[i] + return result @@ -46,7 +46,7 @@ def letterCombinations(self, digits): "pqrs", "tuv", "wxyz"], [] self.letterCombinationsRecu(result, digits, lookup, "", 0) return result - + def letterCombinationsRecu(self, result, digits, lookup, cur, n): if n == len(digits): result.append(cur) diff --git a/Python/lexicographical-numbers.py b/Python/lexicographical-numbers.py index 789aa4d51..20e3f896a 100644 --- a/Python/lexicographical-numbers.py +++ b/Python/lexicographical-numbers.py @@ -18,7 +18,7 @@ def lexicalOrder(self, n): while i * 10**k <= n: result.append(i * 10**k) k += 1 - + num = result[-1] + 1 while num <= n and num % 10: result.append(num) @@ -31,7 +31,7 @@ def lexicalOrder(self, n): while num % 10 == 9: num /= 10 - + i = num+1 return result diff --git a/Python/linked-list-cycle-ii.py b/Python/linked-list-cycle-ii.py index 74ea144ca..281e465fc 100644 --- a/Python/linked-list-cycle-ii.py +++ b/Python/linked-list-cycle-ii.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a linked list, return the node where the cycle begins. If there is no cycle, return null. -# +# # Follow up: # Can you solve it without using extra space? # @@ -12,13 +12,13 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __str__(self): if self: return "{}".format(self.val) else: return None - + class Solution: # @param head, a ListNode # @return a list node diff --git a/Python/linked-list-cycle.py b/Python/linked-list-cycle.py index b2288624d..6073a78be 100644 --- a/Python/linked-list-cycle.py +++ b/Python/linked-list-cycle.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a linked list, determine if it has a cycle in it. -# +# # Follow up: # Can you solve it without using extra space? # diff --git a/Python/linked-list-random-node.py b/Python/linked-list-random-node.py index 1e6d7de0f..7ac0718fc 100644 --- a/Python/linked-list-random-node.py +++ b/Python/linked-list-random-node.py @@ -46,7 +46,7 @@ def getRandom(self): reservoir = curr.val if randint(1, n+1) == 1 else reservoir curr, n = curr.next, n+1 return reservoir - + # Your Solution object will be instantiated and called as such: # obj = Solution(head) diff --git a/Python/longest-absolute-file-path.py b/Python/longest-absolute-file-path.py index bc29abb10..065458aa2 100644 --- a/Python/longest-absolute-file-path.py +++ b/Python/longest-absolute-file-path.py @@ -28,7 +28,7 @@ # For example, in the second example above, the longest absolute path is "dir/subdir2/subsubdir2/file2.ext", # and its length is 32 (not including the double quotes). # -# Given a string representing the file system in the above format, +# Given a string representing the file system in the above format, # return the length of the longest absolute path to file in the abstracted file system. # If there is no file in the system, return 0. # diff --git a/Python/longest-consecutive-sequence.py b/Python/longest-consecutive-sequence.py index df70be6e9..240b109e0 100644 --- a/Python/longest-consecutive-sequence.py +++ b/Python/longest-consecutive-sequence.py @@ -2,11 +2,11 @@ # Space: O(n) # # Given an unsorted array of integers, find the length of the longest consecutive elements sequence. -# +# # For example, # Given [100, 4, 200, 1, 3, 2], # The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. -# +# # Your algorithm should run in O(n) complexity. # diff --git a/Python/longest-continuous-increasing-subsequence.py b/Python/longest-continuous-increasing-subsequence.py index c374bbea3..9ae2635e1 100644 --- a/Python/longest-continuous-increasing-subsequence.py +++ b/Python/longest-continuous-increasing-subsequence.py @@ -7,13 +7,13 @@ # Example 1: # Input: [1,3,5,4,7] # Output: 3 -# Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. +# Explanation: The longest continuous increasing subsequence is [1,3,5], its length is 3. # Even though [1,3,5,7] is also an increasing subsequence, -# it's not a continuous one where 5 and 7 are separated by 4. +# it's not a continuous one where 5 and 7 are separated by 4. # Example 2: # Input: [2,2,2,2,2] # Output: 1 -# Explanation: The longest continuous increasing subsequence is [2], its length is 1. +# Explanation: The longest continuous increasing subsequence is [2], its length is 1. # Note: Length of the array will not exceed 10,000. class Solution(object): diff --git a/Python/longest-increasing-path-in-a-matrix.py b/Python/longest-increasing-path-in-a-matrix.py index f5685d3a7..c95688f2e 100644 --- a/Python/longest-increasing-path-in-a-matrix.py +++ b/Python/longest-increasing-path-in-a-matrix.py @@ -3,8 +3,8 @@ # Given an integer matrix, find the length of the longest increasing path. # -# From each cell, you can either move to four directions: left, right, up -# or down. You may NOT move diagonally or move outside of the boundary +# From each cell, you can either move to four directions: left, right, up +# or down. You may NOT move diagonally or move outside of the boundary # (i.e. wrap-around is not allowed). # # Example 1: @@ -40,7 +40,7 @@ def longestIncreasingPath(self, matrix): def longestpath(matrix, i, j, max_lengths): if max_lengths[i][j]: return max_lengths[i][j] - + max_depth = 0 directions = [(0, -1), (0, 1), (-1, 0), (1, 0)] for d in directions: @@ -56,5 +56,5 @@ def longestpath(matrix, i, j, max_lengths): for i in xrange(len(matrix)): for j in xrange(len(matrix[0])): res = max(res, longestpath(matrix, i, j, max_lengths)) - + return res diff --git a/Python/longest-increasing-subsequence.py b/Python/longest-increasing-subsequence.py index ac477873b..f176202be 100644 --- a/Python/longest-increasing-subsequence.py +++ b/Python/longest-increasing-subsequence.py @@ -1,7 +1,7 @@ # Time: O(nlogn) # Space: O(n) # -# Given an unsorted array of integers, +# Given an unsorted array of integers, # find the length of longest increasing subsequence. # # For example, diff --git a/Python/longest-palindromic-substring.py b/Python/longest-palindromic-substring.py index 2c0f238d9..2621a208d 100644 --- a/Python/longest-palindromic-substring.py +++ b/Python/longest-palindromic-substring.py @@ -24,7 +24,7 @@ def preProcess(s): return T T = preProcess(s) - P = [0] * len(T) + P = [0] * len(T) center, right = 0, 0 for i in xrange(1, len(T) - 1): i_mirror = 2 * center - i @@ -37,8 +37,8 @@ def preProcess(s): P[i] += 1 if i + P[i] > right: - center, right = i, i + P[i] - + center, right = i, i + P[i] + max_i = 0 for i in xrange(1, len(T) - 1): if P[i] > P[max_i]: @@ -46,6 +46,6 @@ def preProcess(s): start = (max_i - 1 - P[max_i]) / 2 return s[start : start + P[max_i]] - + if __name__ == "__main__": print Solution().longestPalindrome("abb") diff --git a/Python/longest-substring-with-at-most-k-distinct-characters.py b/Python/longest-substring-with-at-most-k-distinct-characters.py index a5f6d0c45..fd974b83c 100644 --- a/Python/longest-substring-with-at-most-k-distinct-characters.py +++ b/Python/longest-substring-with-at-most-k-distinct-characters.py @@ -13,12 +13,12 @@ def lengthOfLongestSubstringKDistinct(self, s, k): if visited[ord(char)] == 0: distinct_count += 1 visited[ord(char)] += 1 - + while distinct_count > k: visited[ord(s[start])] -= 1 if visited[ord(s[start])] == 0: distinct_count -= 1 start += 1 - + longest = max(longest, i - start + 1) return longest diff --git a/Python/longest-substring-with-at-most-two-distinct-characters.py b/Python/longest-substring-with-at-most-two-distinct-characters.py index 3a58b3024..0eb42603b 100644 --- a/Python/longest-substring-with-at-most-two-distinct-characters.py +++ b/Python/longest-substring-with-at-most-two-distinct-characters.py @@ -1,11 +1,11 @@ # Time: O(n) # Space: O(1) # -# Given a string, find the length of the longest substring T +# Given a string, find the length of the longest substring T # that contains at most 2 distinct characters. -# +# # For example, Given s = "eceba", -# +# # T is "ece" which its length is 3. # @@ -18,15 +18,15 @@ def lengthOfLongestSubstringTwoDistinct(self, s): if visited[ord(char)] == 0: distinct_count += 1 visited[ord(char)] += 1 - + while distinct_count > 2: visited[ord(s[start])] -= 1 if visited[ord(s[start])] == 0: distinct_count -= 1 start += 1 - + longest = max(longest, i - start + 1) return longest - + if __name__ == "__main__": print Solution().lengthOfLongestSubstringTwoDistinct("eceba") diff --git a/Python/longest-substring-without-repeating-characters.py b/Python/longest-substring-without-repeating-characters.py index c7f711c73..8208f643c 100644 --- a/Python/longest-substring-without-repeating-characters.py +++ b/Python/longest-substring-without-repeating-characters.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a string, find the length of the longest substring without repeating characters. -# For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. +# For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. # For "bbbbb" the longest substring is "b", with the length of 1. # diff --git a/Python/longest-uncommon-subsequence-i.py b/Python/longest-uncommon-subsequence-i.py index 007cee665..9d64937f1 100644 --- a/Python/longest-uncommon-subsequence-i.py +++ b/Python/longest-uncommon-subsequence-i.py @@ -15,9 +15,9 @@ # Example 1: # Input: "aba", "cdc" # Output: 3 -# Explanation: The longest uncommon subsequence is "aba" (or "cdc"), -# because "aba" is a subsequence of "aba", -# but not a subsequence of any other strings in the group of two strings. +# Explanation: The longest uncommon subsequence is "aba" (or "cdc"), +# because "aba" is a subsequence of "aba", +# but not a subsequence of any other strings in the group of two strings. # Note: # # Both strings' lengths will not exceed 100. diff --git a/Python/longest-univalue-path.py b/Python/longest-univalue-path.py index 795f4d485..d7fd3b9b9 100644 --- a/Python/longest-univalue-path.py +++ b/Python/longest-univalue-path.py @@ -52,6 +52,6 @@ def dfs(node): right = (right+1) if node.right and node.right.val == node.val else 0 result[0] = max(result[0], left+right) return max(left, right) - + dfs(root) return result[0] diff --git a/Python/longest-valid-parentheses.py b/Python/longest-valid-parentheses.py index dd0376eb8..3c8888175 100644 --- a/Python/longest-valid-parentheses.py +++ b/Python/longest-valid-parentheses.py @@ -1,11 +1,11 @@ # Time: O(n) # Space: O(1) # -# Given a string containing just the characters '(' and ')', +# Given a string containing just the characters '(' and ')', # find the length of the longest valid (well-formed) parentheses substring. -# +# # For "(()", the longest valid parentheses substring is "()", which has length = 2. -# +# # Another example is ")()())", where the longest valid parentheses substring is "()()", which has length = 4. # diff --git a/Python/longest-word-in-dictionary-through-deleting.py b/Python/longest-word-in-dictionary-through-deleting.py index 7437ca338..548d3797c 100644 --- a/Python/longest-word-in-dictionary-through-deleting.py +++ b/Python/longest-word-in-dictionary-through-deleting.py @@ -12,13 +12,13 @@ # Input: # s = "abpcplea", d = ["ale","apple","monkey","plea"] # -# Output: +# Output: # "apple" # Example 2: # Input: # s = "abpcplea", d = ["a","b","c"] # -# Output: +# Output: # "a" # Note: # All the strings in the input will only contain lower-case letters. diff --git a/Python/lru-cache.py b/Python/lru-cache.py index 14cc5333d..9913918fc 100644 --- a/Python/lru-cache.py +++ b/Python/lru-cache.py @@ -37,7 +37,7 @@ class LinkedList(object): def __init__(self): self.head = None self.tail = None - + def insert(self, node): node.next, node.prev = None, None # avoid dirty node if self.head is None: @@ -46,7 +46,7 @@ def insert(self, node): self.tail.next = node node.prev = self.tail self.tail = node - + def delete(self, node): if node.prev: node.prev.next = node.next @@ -65,12 +65,12 @@ def __init__(self, capacity): self.list = LinkedList() self.dict = {} self.capacity = capacity - + def _insert(self, key, val): node = ListNode(key, val) self.list.insert(node) self.dict[key] = node - + # @return an integer def get(self, key): @@ -80,7 +80,7 @@ def get(self, key): self._insert(key, val) return val return -1 - + # @param key, an integer # @param value, an integer @@ -92,8 +92,8 @@ def put(self, key, val): del self.dict[self.list.head.key] self.list.delete(self.list.head) self._insert(key, val) - - + + import collections class LRUCache2(object): def __init__(self, capacity): @@ -115,7 +115,7 @@ def put(self, key, value): self.cache.popitem(last=False) self.cache[key] = value - + if __name__ == "__main__": cache = LRUCache(3) cache.set(1, 1) @@ -124,4 +124,4 @@ def put(self, key, value): print cache.get(1) cache.set(4, 4) print cache.get(2) - + diff --git a/Python/majority-element-ii.py b/Python/majority-element-ii.py index 849bde6cb..40f02dee3 100644 --- a/Python/majority-element-ii.py +++ b/Python/majority-element-ii.py @@ -1,8 +1,8 @@ # Time: O(n) # Space: O(1) -# Given an integer array of size n, -# find all elements that appear more than [n/3] times. +# Given an integer array of size n, +# find all elements that appear more than [n/3] times. # The algorithm should run in linear time and in O(1) space. import collections diff --git a/Python/majority-element.py b/Python/majority-element.py index f39f2c0d7..2f39c0403 100644 --- a/Python/majority-element.py +++ b/Python/majority-element.py @@ -3,7 +3,7 @@ # # Given an array of size n, find the majority element. # The majority element is the element that appears more than [n/2] times. -# +# # You may assume that the array is non-empty and the majority element always exist in the array. import collections @@ -15,7 +15,7 @@ def majorityElement(self, nums): :rtype: int """ idx, cnt = 0, 1 - + for i in xrange(1, len(nums)): if nums[idx] == nums[i]: cnt += 1 @@ -24,7 +24,7 @@ def majorityElement(self, nums): if cnt == 0: idx = i cnt = 1 - + return nums[idx] def majorityElement2(self, nums): diff --git a/Python/matchsticks-to-square.py b/Python/matchsticks-to-square.py index 9728cccff..18338e2ee 100644 --- a/Python/matchsticks-to-square.py +++ b/Python/matchsticks-to-square.py @@ -44,7 +44,7 @@ def makesquare(self, nums): for subset in xrange(fullset+1): subset_total_len = 0 - for i in xrange(len(nums)): + for i in xrange(len(nums)): if subset & (1 << i): subset_total_len += nums[i] @@ -56,5 +56,5 @@ def makesquare(self, nums): if valid_half_subsets[fullset ^ valid_half_subset]: return True used_subsets.append(subset) - + return False diff --git a/Python/max-chunks-to-make-sorted-ii.py b/Python/max-chunks-to-make-sorted-ii.py index 363d313f9..af0b3e7c7 100644 --- a/Python/max-chunks-to-make-sorted-ii.py +++ b/Python/max-chunks-to-make-sorted-ii.py @@ -39,7 +39,7 @@ def maxChunksToSorted(self, arr): """ def compare(i1, i2): return arr[i1]-arr[i2] if arr[i1] != arr[i2] else i1-i2 - + idxs = [i for i in xrange(len(arr))] result, max_i = 0, 0 for i, v in enumerate(sorted(idxs, cmp=compare)): diff --git a/Python/max-sum-of-sub-matrix-no-larger-than-k.py b/Python/max-sum-of-sub-matrix-no-larger-than-k.py index d21b3fe78..5d18781af 100644 --- a/Python/max-sum-of-sub-matrix-no-larger-than-k.py +++ b/Python/max-sum-of-sub-matrix-no-larger-than-k.py @@ -38,7 +38,7 @@ # Time: O(min(m, n)^2 * max(m, n)^2) # Space: O(max(m, n)) -from bisect import bisect_left, insort +from bisect import bisect_left, insort class Solution(object): def maxSumSubmatrix(self, matrix, k): @@ -59,7 +59,7 @@ def maxSumSubmatrix(self, matrix, k): for j in xrange(i, m): for l in xrange(n): sums[l] += matrix[j][l] if m == len(matrix) else matrix[l][j] - + # Find the max subarray no more than K. accu_sum_set, accu_sum = [0], 0 for sum in sums: @@ -86,7 +86,7 @@ def __init__(self, val): self.val = val self.left = None self.right = None - + def insert(self, val): # Time: O(h) = O(logn) ~ O(n) curr = self while curr: @@ -102,7 +102,7 @@ def insert(self, val): # Time: O(h) = O(logn) ~ O(n) else: curr.right = BST(val) return - + def lower_bound(self, val): # Time: O(h) = O(logn) ~ O(n) result, curr = None, self while curr: @@ -125,7 +125,7 @@ def lower_bound(self, val): # Time: O(h) = O(logn) ~ O(n) for j in xrange(i, m): for l in xrange(n): sums[l] += matrix[j][l] if m == len(matrix) else matrix[l][j] - + # Find the max subarray no more than K. accu_sum_set = BST(0) accu_sum = 0 diff --git a/Python/maximal-rectangle.py b/Python/maximal-rectangle.py index 85420c608..f4f7f374f 100644 --- a/Python/maximal-rectangle.py +++ b/Python/maximal-rectangle.py @@ -1,7 +1,7 @@ # Time: O(n^2) # Space: O(n) -# Given a 2D binary matrix filled with 0's and 1's, +# Given a 2D binary matrix filled with 0's and 1's, # find the largest rectangle containing all ones and return its area. # Ascending stack solution. @@ -49,7 +49,7 @@ def maximalRectangle(self, matrix): """ if not matrix: return 0 - + result = 0 m = len(matrix) n = len(matrix[0]) @@ -68,7 +68,7 @@ def maximalRectangle(self, matrix): H[j] = 0 R[j] = n left = j + 1 - + right = n for j in reversed(xrange(n)): if matrix[i][j] == '1': @@ -76,7 +76,7 @@ def maximalRectangle(self, matrix): result = max(result, H[j] * (R[j] - L[j])) else: right = j - + return result if __name__ == "__main__": diff --git a/Python/maximal-square.py b/Python/maximal-square.py index 2f95f9b99..bb9029526 100644 --- a/Python/maximal-square.py +++ b/Python/maximal-square.py @@ -1,7 +1,7 @@ # Time: O(n^2) # Space: O(n) # -# Given a 2D binary matrix filled with 0's and 1's, +# Given a 2D binary matrix filled with 0's and 1's, # find the largest square containing all 1's and return its area. # # For example, given the following matrix: @@ -24,12 +24,12 @@ def maximalSquare(self, matrix): m, n = len(matrix), len(matrix[0]) size = [[0 for j in xrange(n)] for i in xrange(2)] max_size = 0 - + for j in xrange(n): if matrix[0][j] == '1': size[0][j] = 1 max_size = max(max_size, size[0][j]) - + for i in xrange(1, m): if matrix[i][0] == '1': size[i % 2][0] = 1 @@ -43,7 +43,7 @@ def maximalSquare(self, matrix): max_size = max(max_size, size[i % 2][j]) else: size[i % 2][j] = 0 - + return max_size * max_size @@ -60,12 +60,12 @@ def maximalSquare(self, matrix): m, n = len(matrix), len(matrix[0]) size = [[0 for j in xrange(n)] for i in xrange(m)] max_size = 0 - + for j in xrange(n): if matrix[0][j] == '1': size[0][j] = 1 max_size = max(max_size, size[0][j]) - + for i in xrange(1, m): if matrix[i][0] == '1': size[i][0] = 1 @@ -79,10 +79,10 @@ def maximalSquare(self, matrix): max_size = max(max_size, size[i][j]) else: size[i][j] = 0 - + return max_size * max_size - - + + # Time: O(n^2) # Space: O(n^2) # DP. @@ -92,7 +92,7 @@ class Solution3: def maximalSquare(self, matrix): if not matrix: return 0 - + H, W = 0, 1 # DP table stores (h, w) for each (i, j). table = [[[0, 0] for j in xrange(len(matrix[0]))] \ @@ -108,7 +108,7 @@ def maximalSquare(self, matrix): if j + 1 < len(matrix[i]): w = table[i][j + 1][W] + 1 table[i][j] = [h, w] - + # A table stores the length of largest square for each (i, j). s = [[0 for j in xrange(len(matrix[0]))] \ for i in xrange(len(matrix))] @@ -122,6 +122,6 @@ def maximalSquare(self, matrix): side = min(s[i + 1][j + 1] + 1, side) s[i][j] = side max_square_area = max(max_square_area, side * side) - + return max_square_area; - + diff --git a/Python/maximum-binary-tree.py b/Python/maximum-binary-tree.py index e32134c97..c95d5ff82 100644 --- a/Python/maximum-binary-tree.py +++ b/Python/maximum-binary-tree.py @@ -16,8 +16,8 @@ # 6 # / \ # 3 5 -# \ / -# 2 0 +# \ / +# 2 0 # \ # 1 # Note: diff --git a/Python/maximum-depth-of-binary-tree.py b/Python/maximum-depth-of-binary-tree.py index 016f720a6..2889d824a 100644 --- a/Python/maximum-depth-of-binary-tree.py +++ b/Python/maximum-depth-of-binary-tree.py @@ -2,7 +2,7 @@ # Space: O(h), h is height of binary tree # # Given a binary tree, find its maximum depth. -# +# # The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node. # @@ -19,7 +19,7 @@ class Solution: def maxDepth(self, root): if root is None: return 0 - else: + else: return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1 if __name__ == "__main__": diff --git a/Python/maximum-gap.py b/Python/maximum-gap.py index 2003ab711..06d771637 100644 --- a/Python/maximum-gap.py +++ b/Python/maximum-gap.py @@ -1,16 +1,16 @@ # Time: O(n) # Space: O(n) -# Given an unsorted array, find the maximum difference between -# +# Given an unsorted array, find the maximum difference between +# # the successive elements in its sorted form. -# +# # Try to solve it in linear time/space. -# +# # Return 0 if the array contains less than 2 elements. -# +# # You may assume all elements in the array are non-negative integers -# +# # and fit in the 32-bit signed integer range. # bucket sort @@ -24,7 +24,7 @@ def maximumGap(self, nums): """ if len(nums) < 2: return 0 - + # Init bucket. max_val, min_val = max(nums), min(nums) gap = max(1, (max_val - min_val) / (len(nums) - 1)) @@ -36,11 +36,11 @@ def maximumGap(self, nums): for n in nums: # min_val / max_val is in the first / last bucket. if n in (max_val, min_val): - continue + continue i = (n - min_val) / gap bucket[i]['min'] = min(bucket[i]['min'], n) bucket[i]['max'] = max(bucket[i]['max'], n) - + # Count each bucket gap between the first and the last bucket. max_gap, pre_bucket_max = 0, min_val for i in xrange(bucket_size): @@ -51,8 +51,8 @@ def maximumGap(self, nums): max_gap = max(max_gap, bucket[i]['min'] - pre_bucket_max) pre_bucket_max = bucket[i]['max'] # Count the last bucket. - max_gap = max(max_gap, max_val - pre_bucket_max) - + max_gap = max(max_gap, max_val - pre_bucket_max) + return max_gap @@ -67,11 +67,11 @@ def maximumGap(self, nums): if len(nums) < 2: return 0 - + nums.sort() pre = nums[0] max_gap = float("-inf") - + for i in nums: max_gap = max(max_gap, i - pre) pre = i diff --git a/Python/maximum-product-of-word-lengths.py b/Python/maximum-product-of-word-lengths.py index ddc03bd4b..66a870888 100644 --- a/Python/maximum-product-of-word-lengths.py +++ b/Python/maximum-product-of-word-lengths.py @@ -2,7 +2,7 @@ # Space: O(n) # Given a string array words, find the maximum value of -# length(word[i]) * length(word[j]) where the two words +# length(word[i]) * length(word[j]) where the two words # do not share common letters. You may assume that each # word will contain only lower case letters. If no such # two words exist, return 0. diff --git a/Python/maximum-product-subarray.py b/Python/maximum-product-subarray.py index 50d08388a..cf502bd5f 100644 --- a/Python/maximum-product-subarray.py +++ b/Python/maximum-product-subarray.py @@ -2,7 +2,7 @@ # Space: O(1) # # Find the contiguous subarray within an array (containing at least one number) which has the largest product. -# +# # For example, given the array [2,3,-2,4], # the contiguous subarray [2,3] has the largest product = 6. diff --git a/Python/maximum-subarray.py b/Python/maximum-subarray.py index 35fcd74ae..07a0fcafe 100644 --- a/Python/maximum-subarray.py +++ b/Python/maximum-subarray.py @@ -2,12 +2,12 @@ # Space: O(1) # # Find the contiguous subarray within an array (containing at least one number) which has the largest sum. -# +# # For example, given the array [-2,1,-3,4,-1,2,1,-5,4], # the contiguous subarray [4,-1,2,1] has the largest sum = 6. -# +# # click to show more practice. -# +# # More practice: # If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle. # diff --git a/Python/maximum-sum-of-3-non-overlapping-subarrays.py b/Python/maximum-sum-of-3-non-overlapping-subarrays.py index 7defc2939..30b4f8a1d 100644 --- a/Python/maximum-sum-of-3-non-overlapping-subarrays.py +++ b/Python/maximum-sum-of-3-non-overlapping-subarrays.py @@ -11,7 +11,7 @@ # Example: # Input: [1,2,1,2,6,7,5,1], 2 # Output: [0, 3, 5] - # + # # Explanation: Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5]. # We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger. # @@ -31,7 +31,7 @@ def maxSumOfThreeSubarrays(self, nums, k): accu = [0] for num in nums: accu.append(accu[-1]+num) - + left_pos = [0] * n total = accu[k]-accu[0] for i in xrange(k, n): @@ -40,7 +40,7 @@ def maxSumOfThreeSubarrays(self, nums, k): total = accu[i+1]-accu[i+1-k] else: left_pos[i] = left_pos[i-1] - + right_pos = [n-k] * n total = accu[n]-accu[n-k] for i in reversed(xrange(n-k)): @@ -49,7 +49,7 @@ def maxSumOfThreeSubarrays(self, nums, k): total = accu[i+k]-accu[i] else: right_pos[i] = right_pos[i+1] - + result, max_sum = [], 0 for i in xrange(k, n-2*k+1): left, right = left_pos[i-1], right_pos[i+k] diff --git a/Python/maximum-width-of-binary-tree.py b/Python/maximum-width-of-binary-tree.py index 4bb5125c0..b0362eee8 100644 --- a/Python/maximum-width-of-binary-tree.py +++ b/Python/maximum-width-of-binary-tree.py @@ -10,53 +10,53 @@ # where the null nodes between the end-nodes are also counted into the length calculation. # # Example 1: -# Input: +# Input: # # 1 # / \ # 3 2 -# / \ \ -# 5 3 9 +# / \ \ +# 5 3 9 # # Output: 4 # Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9). # Example 2: -# Input: +# Input: # # 1 -# / -# 3 -# / \ -# 5 3 +# / +# 3 +# / \ +# 5 3 # # Output: 2 # Explanation: The maximum width existing in the third level with the length 2 (5,3). # Example 3: -# Input: +# Input: # # 1 # / \ -# 3 2 -# / -# 5 +# 3 2 +# / +# 5 # # Output: 2 # Explanation: The maximum width existing in the second level with the length 2 (3,2). # Example 4: -# Input: +# Input: # # 1 # / \ # 3 2 -# / \ -# 5 9 +# / \ +# 5 9 # / \ # 6 7 # Output: 8 # Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7). # # Note: Answer will in the range of 32-bit signed integer. -# +# # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): diff --git a/Python/maximum-xor-of-two-numbers-in-an-array.py b/Python/maximum-xor-of-two-numbers-in-an-array.py index 20f541afe..6a3156b19 100644 --- a/Python/maximum-xor-of-two-numbers-in-an-array.py +++ b/Python/maximum-xor-of-two-numbers-in-an-array.py @@ -22,7 +22,7 @@ def findMaximumXOR(self, nums): :rtype: int """ result = 0 - + for i in reversed(xrange(32)): result <<= 1 prefixes = set() diff --git a/Python/median-of-two-sorted-arrays.py b/Python/median-of-two-sorted-arrays.py index 14ce80108..2450b65bf 100644 --- a/Python/median-of-two-sorted-arrays.py +++ b/Python/median-of-two-sorted-arrays.py @@ -1,6 +1,6 @@ # Time: O(log(min(m, n))) # Space: O(1) - + # There are two sorted arrays nums1 and nums2 of size m and n respectively. # Find the median of the two sorted arrays. # The overall run time complexity should be O(log (m+n)). @@ -13,7 +13,7 @@ def findMedianSortedArrays(self, nums1, nums2): :rtype: float """ len1, len2 = len(nums1), len(nums2) - if (len1 + len2) % 2 == 1: + if (len1 + len2) % 2 == 1: return self.getKth(nums1, nums2, (len1 + len2)/2 + 1) else: return (self.getKth(nums1, nums2, (len1 + len2)/2) + \ @@ -24,7 +24,7 @@ def getKth(self, A, B, k): if m > n: return self.getKth(B, A, k) - left, right = 0, m + left, right = 0, m while left < right: mid = left + (right - left) / 2 if 0 <= k - 1 - mid < n and A[mid] >= B[k - 1 - mid]: @@ -49,7 +49,7 @@ def findMedianSortedArrays(self, nums1, nums2): :rtype: float """ len1, len2 = len(nums1), len(nums2) - if (len1 + len2) % 2 == 1: + if (len1 + len2) % 2 == 1: return self.getKth([nums1, nums2], (len1 + len2)/2 + 1) else: return (self.getKth([nums1, nums2], (len1 + len2)/2) + \ @@ -85,4 +85,4 @@ def match(arrays, num, target): if __name__ == "__main__": print Solution().findMedianSortedArrays([1, 3, 5, 7], [2, 4, 6]) print Solution().findMedianSortedArrays([1, 3, 5], [2, 4, 6]) - + diff --git a/Python/meeting-rooms.py b/Python/meeting-rooms.py index f8ef52f66..de7d864c6 100644 --- a/Python/meeting-rooms.py +++ b/Python/meeting-rooms.py @@ -12,7 +12,7 @@ class Solution: # @return {boolean} def canAttendMeetings(self, intervals): intervals.sort(key=lambda x: x.start) - + for i in xrange(1, len(intervals)): if intervals[i].start < intervals[i-1].end: return False diff --git a/Python/merge-intervals.py b/Python/merge-intervals.py index 953798275..1e279e4d6 100644 --- a/Python/merge-intervals.py +++ b/Python/merge-intervals.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a collection of intervals, merge all overlapping intervals. -# +# # For example, # Given [1,3],[2,6],[8,10],[15,18], # return [1,6],[8,10],[15,18]. @@ -13,7 +13,7 @@ class Interval: def __init__(self, s=0, e=0): self.start = s self.end = e - + def __repr__(self): return "[{}, {}]".format(self.start, self.end) @@ -30,7 +30,7 @@ def merge(self, intervals): result = [intervals[0]] for i in xrange(1, len(intervals)): prev, current = result[-1], intervals[i] - if current.start <= prev.end: + if current.start <= prev.end: prev.end = max(prev.end, current.end) else: result.append(current) diff --git a/Python/merge-k-sorted-lists.py b/Python/merge-k-sorted-lists.py index edc3c9d4a..0acffe1af 100644 --- a/Python/merge-k-sorted-lists.py +++ b/Python/merge-k-sorted-lists.py @@ -67,7 +67,7 @@ def mergeTwoLists(l1, l2): curr = curr.next curr.next = l1 or l2 return dummy.next - + def mergeKListsHelper(lists, begin, end): if begin > end: return None @@ -75,7 +75,7 @@ def mergeKListsHelper(lists, begin, end): return lists[begin] return mergeTwoLists(mergeKListsHelper(lists, begin, (begin + end) / 2), \ mergeKListsHelper(lists, (begin + end) / 2 + 1, end)) - + return mergeKListsHelper(lists, 0, len(lists) - 1) @@ -89,19 +89,19 @@ class Solution3: def mergeKLists(self, lists): dummy = ListNode(0) current = dummy - + heap = [] for sorted_list in lists: if sorted_list: heapq.heappush(heap, (sorted_list.val, sorted_list)) - + while heap: smallest = heapq.heappop(heap)[1] current.next = smallest current = current.next if smallest.next: heapq.heappush(heap, (smallest.next.val, smallest.next)) - + return dummy.next @@ -110,5 +110,5 @@ def mergeKLists(self, lists): list1.next = ListNode(3) list2 = ListNode(2) list2.next = ListNode(4) - + print Solution().mergeKLists([list1, list2]) diff --git a/Python/merge-sorted-array.py b/Python/merge-sorted-array.py index 5b6a275fc..b8ace9bb5 100644 --- a/Python/merge-sorted-array.py +++ b/Python/merge-sorted-array.py @@ -2,9 +2,9 @@ # Space: O(1) # # Given two sorted integer arrays A and B, merge B into A as one sorted array. -# +# # Note: -# You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. +# You may assume that A has enough space (size that is greater or equal to m + n) to hold additional elements from B. # The number of elements initialized in A and B are m and n respectively. # @@ -16,7 +16,7 @@ class Solution: # @return nothing def merge(self, A, m, B, n): last, i, j = m + n - 1, m - 1, n - 1 - + while i >= 0 and j >= 0: if A[i] > B[j]: A[last] = A[i] @@ -24,7 +24,7 @@ def merge(self, A, m, B, n): else: A[last] = B[j] last, j = last - 1, j - 1 - + while j >= 0: A[last] = B[j] last, j = last - 1, j - 1 @@ -34,7 +34,7 @@ def merge(self, A, m, B, n): B = [2, 4, 6, 7] Solution().merge(A, 3, B, 4) print A - + # Time: O(n) # Space: O(n) diff --git a/Python/merge-two-binary-trees.py b/Python/merge-two-binary-trees.py index 3fc1c04ef..89341b1e3 100644 --- a/Python/merge-two-binary-trees.py +++ b/Python/merge-two-binary-trees.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(h) -# Given two binary trees and imagine that +# Given two binary trees and imagine that # when you put one of them to cover the other, # some nodes of the two trees are overlapped # while the others are not. @@ -12,21 +12,21 @@ # Otherwise, the NOT null node will be used as the node of new tree. # # Example 1: -# Input: -# Tree 1 Tree 2 -# 1 2 -# / \ / \ -# 3 2 1 3 -# / \ \ -# 5 4 7 -# Output: +# Input: +# Tree 1 Tree 2 +# 1 2 +# / \ / \ +# 3 2 1 3 +# / \ \ +# 5 4 7 +# Output: # Merged tree: # 3 # / \ # 4 5 -# / \ \ +# / \ \ # 5 4 7 -# +# # Note: The merging process must start from the root nodes of both trees. # Definition for a binary tree node. diff --git a/Python/merge-two-sorted-lists.py b/Python/merge-two-sorted-lists.py index 0d965d2b9..c94df5a8b 100644 --- a/Python/merge-two-sorted-lists.py +++ b/Python/merge-two-sorted-lists.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(1) # -# Merge two sorted linked lists and return it as a new list. +# Merge two sorted linked lists and return it as a new list. # The new list should be made by splicing together the nodes of the first two lists. # @@ -42,5 +42,5 @@ def mergeTwoLists(self, l1, l2): l2 = ListNode (2) l2.next = ListNode(3) print Solution().mergeTwoLists(l1, l2) - - + + diff --git a/Python/min-stack.py b/Python/min-stack.py index 12df16163..ff31748a8 100644 --- a/Python/min-stack.py +++ b/Python/min-stack.py @@ -2,7 +2,7 @@ # Space: O(1) # # Design a stack that supports push, pop, top, and retrieving the minimum element in constant time. -# +# # push(x) -- Push element x onto stack. # pop() -- Removes the element on top of the stack. # top() -- Get the top element. @@ -13,7 +13,7 @@ class MinStack: def __init__(self): self.min = None self.stack = [] - + # @param x, an integer # @return an integer def push(self, x): @@ -38,7 +38,7 @@ def top(self): return x + self.min else: return self.min - + # @return an integer def getMin(self): return self.min @@ -67,11 +67,11 @@ def pop(self): self.minStack[-1][1] -= 1 if self.minStack[-1][1] == 0: self.minStack.pop() - + # @return an integer def top(self): return self.stack[-1] - + # @return an integer def getMin(self): return self.minStack[-1][0] @@ -80,4 +80,4 @@ def getMin(self): stack = MinStack() stack.push(-1) print [stack.top(), stack.getMin()] - + diff --git a/Python/minimize-max-distance-to-gas-station.py b/Python/minimize-max-distance-to-gas-station.py index 4b56a68e8..a8676a4a4 100644 --- a/Python/minimize-max-distance-to-gas-station.py +++ b/Python/minimize-max-distance-to-gas-station.py @@ -13,7 +13,7 @@ # # Input: stations = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], K = 9 # Output: 0.500000 -# +# # Note: # - stations.length will be an integer in range [10, 2000]. # - stations[i] will be an integer in range [0, 10^8]. diff --git a/Python/minimum-depth-of-binary-tree.py b/Python/minimum-depth-of-binary-tree.py index 2833dcaab..709093e51 100644 --- a/Python/minimum-depth-of-binary-tree.py +++ b/Python/minimum-depth-of-binary-tree.py @@ -2,7 +2,7 @@ # Space: O(h), h is height of binary tree # # Given a binary tree, find its minimum depth. -# +# # The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node. # @@ -19,7 +19,7 @@ class Solution: def minDepth(self, root): if root is None: return 0 - + if root.left and root.right: return min(self.minDepth(root.left), self.minDepth(root.right)) + 1 else: diff --git a/Python/minimum-distance-between-bst-nodes.py b/Python/minimum-distance-between-bst-nodes.py index ca26974c7..dfb2ad825 100644 --- a/Python/minimum-distance-between-bst-nodes.py +++ b/Python/minimum-distance-between-bst-nodes.py @@ -16,8 +16,8 @@ # 4 # / \ # 2 6 -# / \ -# 1 3 +# / \ +# 1 3 # # while the minimum difference in this tree is 1, # it occurs between node 1 and node 2, also between node 3 and node 2. @@ -51,4 +51,4 @@ def dfs(node): self.result = float('inf') dfs(root) return self.result - + diff --git a/Python/minimum-factorization.py b/Python/minimum-factorization.py index 52d72293a..756c20e81 100644 --- a/Python/minimum-factorization.py +++ b/Python/minimum-factorization.py @@ -16,4 +16,4 @@ def smallestFactorization(self, a): result = mul*i + result mul *= 10 return result if a == 1 and result < 2**31 else 0 - + diff --git a/Python/minimum-genetic-mutation.py b/Python/minimum-genetic-mutation.py index 900510bb7..bd89371fe 100644 --- a/Python/minimum-genetic-mutation.py +++ b/Python/minimum-genetic-mutation.py @@ -1,26 +1,26 @@ # Time: O(n * b), n is the length of gene string, b is size of bank # Space: O(b) -# A gene string can be represented by an 8-character long string, -# with choices from "A","C","G","T". -# Suppose we need to investigate about a mutation (mutation from "start" to "end"), -# where ONE mutation is defined as ONE single character changed in the gene string. -# For example, "AACCGGTT" -> "AACCGGTA" is 1 mutation. +# A gene string can be represented by an 8-character long string, +# with choices from "A","C","G","T". +# Suppose we need to investigate about a mutation (mutation from "start" to "end"), +# where ONE mutation is defined as ONE single character changed in the gene string. +# For example, "AACCGGTT" -> "AACCGGTA" is 1 mutation. # Also, there is a given gene "bank", which records all the valid gene mutations. -# A gene must be in the bank to make it a valid gene string. +# A gene must be in the bank to make it a valid gene string. # # Now, given 3 things - start, end, bank, # your task is to determine what is the minimum number of mutations needed to -# mutate from "start" to "end". If there is no such a mutation, return -1. +# mutate from "start" to "end". If there is no such a mutation, return -1. # # NOTE: 1. Starting point is assumed to be valid, so it might not be included in the bank. # 2. If multiple mutations are needed, all mutations during in the sequence must be valid. # -# For example, -# -# bank: "AACCGGTA" -# start: "AACCGGTT" -# end: "AACCGGTA" +# For example, +# +# bank: "AACCGGTA" +# start: "AACCGGTT" +# end: "AACCGGTA" # return: 1 # # bank: "AACCGGTA", "AACCGCTA", "AAACGGTA" diff --git a/Python/minimum-moves-to-equal-array-elements-ii.py b/Python/minimum-moves-to-equal-array-elements-ii.py index ee447cd6f..49207b3bd 100644 --- a/Python/minimum-moves-to-equal-array-elements-ii.py +++ b/Python/minimum-moves-to-equal-array-elements-ii.py @@ -19,7 +19,7 @@ def PartitionAroundPivot(left, right, pivot_idx, nums): if nums[i] > pivot_value: nums[i], nums[new_pivot_idx] = nums[new_pivot_idx], nums[i] new_pivot_idx += 1 - + nums[right], nums[new_pivot_idx] = nums[new_pivot_idx], nums[right] return new_pivot_idx diff --git a/Python/minimum-path-sum.py b/Python/minimum-path-sum.py index f6c4edfd0..af07387ca 100644 --- a/Python/minimum-path-sum.py +++ b/Python/minimum-path-sum.py @@ -1,9 +1,9 @@ # Time: O(m * n) # Space: O(m + n) # -# Given a m x n grid filled with non-negative numbers, +# Given a m x n grid filled with non-negative numbers, # find a path from top left to bottom right which minimizes the sum of all numbers along its path. -# +# # Note: You can only move either down or right at any point in time. # @@ -19,9 +19,9 @@ def minPathSum(self, grid): sum[0] += grid[i][0] for j in xrange(1, len(grid[0])): sum[j] = min(sum[j - 1], sum[j]) + grid[i][j] - + return sum[-1] - + if __name__ == "__main__": print Solution().minPathSum([[0,1] ,[1,0]]) diff --git a/Python/minimum-size-subarray-sum.py b/Python/minimum-size-subarray-sum.py index 24d56ced0..e9ba17541 100644 --- a/Python/minimum-size-subarray-sum.py +++ b/Python/minimum-size-subarray-sum.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(1) # -# Given an array of n positive integers and a positive integer s, +# Given an array of n positive integers and a positive integer s, # find the minimal length of a subarray of which the sum ≥ s. If there isn't one, return 0 instead. # # For example, given the array [2,3,1,2,4,3] and s = 7, @@ -49,7 +49,7 @@ def minSubArrayLen(self, s, nums): min_size = min(min_size, end - i + 1) return min_size if min_size != float("inf") else 0 - + def binarySearch(self, compare, A, start, end, target): while start < end: mid = start + (end - start) / 2 diff --git a/Python/minimum-swaps-to-make-sequences-increasing.py b/Python/minimum-swaps-to-make-sequences-increasing.py index d3806d0d9..4ba4acc85 100644 --- a/Python/minimum-swaps-to-make-sequences-increasing.py +++ b/Python/minimum-swaps-to-make-sequences-increasing.py @@ -15,7 +15,7 @@ # Example: # Input: A = [1,3,5,4], B = [1,2,3,7] # Output: 1 -# Explanation: +# Explanation: # Swap A[3] and B[3]. Then the sequences are: # A = [1, 3, 5, 7] and B = [1, 2, 3, 4] # which are both strictly increasing. diff --git a/Python/minimum-window-subsequence.py b/Python/minimum-window-subsequence.py index 647dcea4c..3fb65630c 100644 --- a/Python/minimum-window-subsequence.py +++ b/Python/minimum-window-subsequence.py @@ -12,7 +12,7 @@ def minWindow(self, S, T): for j, c in enumerate(S): if c == T[0]: dp[0][j] = j - + for i in xrange(1, len(T)): prev = None dp[i%2] = [None] * len(S) diff --git a/Python/minimum-window-substring.py b/Python/minimum-window-substring.py index d006f4f1e..d70f8c212 100644 --- a/Python/minimum-window-substring.py +++ b/Python/minimum-window-substring.py @@ -3,16 +3,16 @@ # Given a string S and a string T, find the minimum window in S which # will contain all the characters in T in complexity O(n). -# +# # For example, # S = "ADOBECODEBANC" # T = "ABC" # Minimum window is "BANC". -# +# # Note: # If there is no such window in S that covers all characters in T, # return the emtpy string "". -# +# # If there are multiple such windows, you are guaranteed that # there will always be only one unique minimum window in S. @@ -25,32 +25,32 @@ def minWindow(self, s, t): """ current_count = [0 for i in xrange(52)] expected_count = [0 for i in xrange(52)] - + for char in t: expected_count[ord(char) - ord('a')] += 1 - + i, count, start, min_width, min_start = 0, 0, 0, float("inf"), 0 while i < len(s): current_count[ord(s[i]) - ord('a')] += 1 if current_count[ord(s[i]) - ord('a')] <= expected_count[ord(s[i]) - ord('a')]: count += 1 - + if count == len(t): while expected_count[ord(s[start]) - ord('a')] == 0 or \ current_count[ord(s[start]) - ord('a')] > expected_count[ord(s[start]) - ord('a')]: current_count[ord(s[start]) - ord('a')] -= 1 start += 1 - + if min_width > i - start + 1: min_width = i - start + 1 min_start = start i += 1 - + if min_width == float("inf"): return "" - + return s[min_start:min_start + min_width] if __name__ == "__main__": - print Solution().minWindow("ADOBECODEBANC", "ABC") + print Solution().minWindow("ADOBECODEBANC", "ABC") diff --git a/Python/missing-ranges.py b/Python/missing-ranges.py index e43b6104c..16f3a36a5 100644 --- a/Python/missing-ranges.py +++ b/Python/missing-ranges.py @@ -3,8 +3,8 @@ # # Given a sorted integer array where the range of elements are [lower, upper] inclusive, # return its missing ranges. -# -# For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, +# +# For example, given [0, 1, 3, 50, 75], lower = 0 and upper = 99, # return ["2", "4->49", "51->74", "76->99"]. # @@ -23,7 +23,7 @@ def getRange(lower, upper): return "{}->{}".format(lower, upper) ranges = [] pre = lower - 1 - + for i in xrange(len(nums) + 1): if i == len(nums): cur = upper + 1 @@ -31,9 +31,9 @@ def getRange(lower, upper): cur = nums[i] if cur - pre >= 2: ranges.append(getRange(pre + 1, cur - 1)) - + pre = cur - + return ranges diff --git a/Python/monotone-increasing-digits.py b/Python/monotone-increasing-digits.py index 111fb299c..b7b1e508f 100644 --- a/Python/monotone-increasing-digits.py +++ b/Python/monotone-increasing-digits.py @@ -14,11 +14,11 @@ # Example 2: # Input: N = 1234 # Output: 1234 -# +# # Example 3: # Input: N = 332 # Output: 299 -# +# # Note: N is an integer in the range [0, 10^9]. class Solution(object): diff --git a/Python/move-zeroes.py b/Python/move-zeroes.py index fb121754c..410eb8ffe 100644 --- a/Python/move-zeroes.py +++ b/Python/move-zeroes.py @@ -2,10 +2,10 @@ # Space: O(1) # Given an array nums, write a function to move all 0's -# to the end of it while maintaining the relative order +# to the end of it while maintaining the relative order # of the non-zero elements. # -# For example, given nums = [0, 1, 0, 3, 12], after +# For example, given nums = [0, 1, 0, 3, 12], after # calling your function, nums should be [1, 3, 12, 0, 0]. # # Note: @@ -44,9 +44,9 @@ def moveZeroes(self, nums): if nums[i]: nums[pos] = nums[i] pos += 1 - + for i in xrange(pos, len(nums)): - nums[i] = 0 + nums[i] = 0 if __name__ == '__main__': diff --git a/Python/multiply-strings.py b/Python/multiply-strings.py index 6df6c3f12..cfb6d12d8 100644 --- a/Python/multiply-strings.py +++ b/Python/multiply-strings.py @@ -2,7 +2,7 @@ # Space: O(m + n) # # Given two numbers represented as strings, return multiplication of the numbers as a string. -# +# # Note: The numbers can be arbitrarily large and are non-negative. # diff --git a/Python/my-calendar-i.py b/Python/my-calendar-i.py index 8814fc510..050d01f9b 100644 --- a/Python/my-calendar-i.py +++ b/Python/my-calendar-i.py @@ -21,7 +21,7 @@ # MyCalendar.book(10, 20); // returns true # MyCalendar.book(15, 25); // returns false # MyCalendar.book(20, 30); // returns true -# Explanation: +# Explanation: # The first event can be booked. The second can't because time 15 is already booked by another event. # The third event can be booked, as the first event takes every time less than 20, but not including 20. # @@ -87,7 +87,7 @@ def book(self, start, end): if start < j and end > i: return False self.__calendar.append((start, end)) - return True + return True # Your MyCalendar object will be instantiated and called as such: diff --git a/Python/my-calendar-ii.py b/Python/my-calendar-ii.py index 0dc4f14e6..104ccf4b5 100644 --- a/Python/my-calendar-ii.py +++ b/Python/my-calendar-ii.py @@ -25,7 +25,7 @@ # MyCalendar.book(5, 10); // returns true # MyCalendar.book(25, 55); // returns true # -# Explanation: +# Explanation: # The first two events can be booked. The third event can be double booked. # The fourth event (5, 15) can't be booked, because it would result in a triple booking. # The fifth event (5, 10) can be booked, as it does not use time 10 which is already double booked. @@ -35,7 +35,7 @@ # Note: # - The number of calls to MyCalendar.book per test case will be at most 1000. # - In calls to MyCalendar.book(start, end), start and end are integers in the range [0, 10^9]. - + class MyCalendarTwo(object): def __init__(self): diff --git a/Python/n-queens-ii.py b/Python/n-queens-ii.py index 56e4c8f22..ee48742e4 100644 --- a/Python/n-queens-ii.py +++ b/Python/n-queens-ii.py @@ -2,7 +2,7 @@ # Space: O(n) # # Follow up for N-Queens problem. -# +# # Now, instead outputting board configurations, return the total number of distinct solutions. # @@ -14,7 +14,7 @@ def totalNQueens(self, n): self.main_diag = [False] * (2 * n) self.anti_diag = [False] * (2 * n) return self.totalNQueensRecu([], 0, n) - + def totalNQueensRecu(self, solution, row, n): if row == n: return 1 @@ -31,7 +31,7 @@ class Solution2: # @return an integer def totalNQueens(self, n): return self.totalNQueensRecu([], 0, n) - + def totalNQueensRecu(self, solution, row, n): if row == n: return 1 diff --git a/Python/n-queens.py b/Python/n-queens.py index 9ca999b96..6140a2e53 100644 --- a/Python/n-queens.py +++ b/Python/n-queens.py @@ -1,23 +1,23 @@ # Time: O(n!) # Space: O(n) -# The n-queens puzzle is the problem of placing n queens on +# The n-queens puzzle is the problem of placing n queens on # an nxn chess board such that no two queens attack each other. -# +# # Given an integer n, return all distinct solutions to the n-queens puzzle. -# -# Each solution contains a distinct board configuration of the n-queens' placement, +# +# Each solution contains a distinct board configuration of the n-queens' placement, # where 'Q' and '.' both indicate a queen and an empty space respectively. -# +# # For example, # There exist two distinct solutions to the 4-queens puzzle: -# +# # [ # [".Q..", // Solution 1 # "...Q", # "Q...", # "..Q."], -# +# # ["..Q.", // Solution 2 # "Q...", # "...Q", @@ -43,22 +43,22 @@ def dfs(curr, cols, main_diag, anti_diag, result): dfs(curr, cols, main_diag, anti_diag, result) curr.pop() cols[i] = main_diag[row+i] = anti_diag[row-i+n] = False - + result = [] cols, main_diag, anti_diag = [False]*n, [False]*(2*n), [False]*(2*n) dfs([], cols, main_diag, anti_diag, result) return result - + # For any point (x,y), if we want the new point (p,q) don't share the same row, column, or diagonal. -# then there must have ```p+q != x+y``` and ```p-q!= x-y``` -# the former focus on eliminate 'left bottom right top' diagonal; +# then there must have ```p+q != x+y``` and ```p-q!= x-y``` +# the former focus on eliminate 'left bottom right top' diagonal; # the latter focus on eliminate 'left top right bottom' diagonal # - col_per_row: the list of column index per row # - cur_row：current row we are seraching for valid column # - xy_diff：the list of x-y -# - xy_sum：the list of x+y +# - xy_sum：the list of x+y class Solution2(object): def solveNQueens(self, n): """ @@ -75,7 +75,7 @@ def dfs(col_per_row, xy_diff, xy_sum): ress = [] dfs([], [], []) return [['.'*i + 'Q' + '.'*(n-i-1) for i in res] for res in ress] - - + + if __name__ == "__main__": print Solution().solveNQueens(8) diff --git a/Python/nested-list-weight-sum-ii.py b/Python/nested-list-weight-sum-ii.py index 0594d091c..8eb003d4f 100644 --- a/Python/nested-list-weight-sum-ii.py +++ b/Python/nested-list-weight-sum-ii.py @@ -40,7 +40,7 @@ def depthSumInverseHelper(list, depth, result): else: for l in list.getList(): depthSumInverseHelper(l, depth + 1, result) - + result = [] for list in nestedList: depthSumInverseHelper(list, 0, result) diff --git a/Python/next-greater-element-ii.py b/Python/next-greater-element-ii.py index 03ee1c106..6a75d5e1f 100644 --- a/Python/next-greater-element-ii.py +++ b/Python/next-greater-element-ii.py @@ -10,8 +10,8 @@ # Example 1: # Input: [1,2,1] # Output: [2,-1,2] -# Explanation: The first 1's next greater number is 2; -# The number 2 can't find next greater number; +# Explanation: The first 1's next greater number is 2; +# The number 2 can't find next greater number; # The second 1's next greater number needs to search circularly, which is also 2. # Note: The length of given array won't exceed 10000. diff --git a/Python/next-permutation.py b/Python/next-permutation.py index 48056e9c6..def9a572d 100644 --- a/Python/next-permutation.py +++ b/Python/next-permutation.py @@ -2,11 +2,11 @@ # Space: O(1) # # Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. -# +# # If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending order). -# +# # The replacement must be in-place, do not allocate extra memory. -# +# # Here are some examples. Inputs are in the left-hand column and its corresponding outputs are in the right-hand column. # 1,2,3 -> 1,3,2 # 3,2,1 -> 1,2,3 @@ -21,15 +21,15 @@ def nextPermutation(self, num): for i in xrange(len(num) - 1): if num[i] < num[i + 1]: k = i - + if k == -1: num.reverse() return - + for i in xrange(k + 1, len(num)): if num[i] > num[k]: l = i - + num[k], num[l] = num[l], num[k] num[k + 1:] = num[:k:-1] diff --git a/Python/nim-game.py b/Python/nim-game.py index f1eadb01a..e14d788df 100644 --- a/Python/nim-game.py +++ b/Python/nim-game.py @@ -2,13 +2,13 @@ # Space: O(1) # # You are playing the following Nim Game with your friend: -# There is a heap of stones on the table, each time one of +# There is a heap of stones on the table, each time one of # you take turns to remove 1 to 3 stones. # The one who removes the last stone will be the winner. # You will take the first turn to remove the stones. # # Both of you are very clever and have optimal strategies for -# the game. Write a function to determine whether you can win +# the game. Write a function to determine whether you can win # the game given the number of stones in the heap. # # For example, if there are 4 stones in the heap, then you will diff --git a/Python/non-decreasing-array.py b/Python/non-decreasing-array.py index e5e51c42f..d556d531d 100644 --- a/Python/non-decreasing-array.py +++ b/Python/non-decreasing-array.py @@ -9,9 +9,9 @@ # Example 1: # Input: [4,2,3] # Output: True -# Explanation: You could modify the first +# Explanation: You could modify the first # 4 -# to +# to # 1 # to get a non-decreasing array. # Example 2: @@ -39,4 +39,4 @@ def checkPossibility(self, nums): else: prev = nums[i] return True - + diff --git a/Python/non-negative-integers-without-consecutive-ones.py b/Python/non-negative-integers-without-consecutive-ones.py index 37d59f055..5847fe8ba 100644 --- a/Python/non-negative-integers-without-consecutive-ones.py +++ b/Python/non-negative-integers-without-consecutive-ones.py @@ -7,7 +7,7 @@ # Example 1: # Input: 5 # Output: 5 -# Explanation: +# Explanation: # Here are the non-negative integers <= 5 with their corresponding binary representations: # 0 : 0 # 1 : 1 @@ -15,7 +15,7 @@ # 3 : 11 # 4 : 100 # 5 : 101 -# Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. +# Among them, only integer 3 disobeys the rule (two consecutive ones) and the other 5 satisfy the rule. # Note: 1 <= n <= 10^9 class Solution(object): diff --git a/Python/number-of-1-bits.py b/Python/number-of-1-bits.py index 1d3f12f2a..065c5db7c 100644 --- a/Python/number-of-1-bits.py +++ b/Python/number-of-1-bits.py @@ -1,10 +1,10 @@ # Time: O(logn) = O(32) # Space: O(1) # -# Write a function that takes an unsigned integer +# Write a function that takes an unsigned integer # and returns the number of '1' bits it has (also known as the Hamming weight). -# -# For example, the 32-bit integer '11' has binary representation 00000000000000000000000000001011, +# +# For example, the 32-bit integer '11' has binary representation 00000000000000000000000000001011, # so the function should return 3. # class Solution: diff --git a/Python/number-of-boomerangs.py b/Python/number-of-boomerangs.py index e31756e0f..84ff01205 100644 --- a/Python/number-of-boomerangs.py +++ b/Python/number-of-boomerangs.py @@ -1,7 +1,7 @@ # Time: O(n^2) # Space: O(n) -# Given n points in the plane that are all pairwise distinct, +# Given n points in the plane that are all pairwise distinct, # a "boomerang" is a tuple of points (i, j, k) such that the distance # between i and j equals the distance between i and k (the order of the tuple matters). # @@ -35,11 +35,11 @@ def numberOfBoomerangs(self, points): continue dx, dy = points[i][0] - points[j][0], points[i][1] - points[j][1] group[dx**2 + dy**2] += 1 - + for _, v in group.iteritems(): if v > 1: result += v * (v-1) - + return result def numberOfBoomerangs2(self, points): diff --git a/Python/number-of-corner-rectangles.py b/Python/number-of-corner-rectangles.py index 6dbe75bee..5988e1885 100644 --- a/Python/number-of-corner-rectangles.py +++ b/Python/number-of-corner-rectangles.py @@ -7,7 +7,7 @@ # Note that only the corners need to have the value 1. Also, all four 1s used must be distinct. # # Example 1: -# Input: grid = +# Input: grid = # [[1, 0, 0, 1, 0], # [0, 0, 1, 0, 1], # [0, 0, 0, 1, 0], @@ -16,7 +16,7 @@ # Explanation: There is only one corner rectangle, with corners grid[1][2], grid[1][4], grid[3][2], grid[3][4]. # # Example 2: -# Input: grid = +# Input: grid = # [[1, 1, 1], # [1, 1, 1], # [1, 1, 1]] @@ -24,7 +24,7 @@ # Explanation: There are four 2x2 rectangles, four 2x3 and 3x2 rectangles, and one 3x3 rectangle. # # Example 3: -# Input: grid = +# Input: grid = # [[1, 1, 1, 1]] # Output: 0 # Explanation: Rectangles must have four distinct corners. diff --git a/Python/number-of-distinct-islands-ii.py b/Python/number-of-distinct-islands-ii.py index 4b1197c5c..49de00acc 100644 --- a/Python/number-of-distinct-islands-ii.py +++ b/Python/number-of-distinct-islands-ii.py @@ -34,7 +34,7 @@ def normalize(island): p[0] -= origin[0] p[1] -= origin[1] return min(shapes) - + islands = set() for i in xrange(len(grid)): for j in xrange(len(grid[0])): diff --git a/Python/number-of-islands-ii.py b/Python/number-of-islands-ii.py index 3c36a265e..c76a1a5df 100644 --- a/Python/number-of-islands-ii.py +++ b/Python/number-of-islands-ii.py @@ -20,7 +20,7 @@ def find_set(x): def union_set(x, y): x_root, y_root = find_set(x), find_set(y) set[min(x_root, y_root)] = max(x_root, y_root) - + numbers = [] number = 0 directions = [(0, -1), (0, 1), (-1, 0), (1, 0)] diff --git a/Python/number-of-islands.py b/Python/number-of-islands.py index 89bff89d1..6f34748dd 100644 --- a/Python/number-of-islands.py +++ b/Python/number-of-islands.py @@ -1,8 +1,8 @@ # Time: O(m * n) # Space: O(m * n) # -# Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. -# An island is surrounded by water and is formed by connecting adjacent lands horizontally +# Given a 2d grid map of '1's (land) and '0's (water), count the number of islands. +# An island is surrounded by water and is formed by connecting adjacent lands horizontally # or vertically. You may assume all four edges of the grid are all surrounded by water. # # Example 1: @@ -28,9 +28,9 @@ class Solution: def numIslands(self, grid): if not grid: return 0 - + row = len(grid) - col = len(grid[0]) + col = len(grid[0]) count = 0 for i in xrange(row): for j in xrange(col): @@ -43,7 +43,7 @@ def dfs(self, grid, row, col, x, y): if grid[x][y] == '0': return grid[x][y] = '0' - + if x != 0: self.dfs(grid, row, col, x - 1, y) if x != row - 1: diff --git a/Python/number-of-lines-to-write-string.py b/Python/number-of-lines-to-write-string.py index 886f139c3..b2e047002 100644 --- a/Python/number-of-lines-to-write-string.py +++ b/Python/number-of-lines-to-write-string.py @@ -11,22 +11,22 @@ # and what is the width used by the last such line? Return your answer as an integer list of length 2. # # Example : -# Input: +# Input: # widths = [10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10] # S = "abcdefghijklmnopqrstuvwxyz" # Output: [3, 60] # -# Explanation: +# Explanation: # All letters have the same length of 10. To write all 26 letters, # we need two full lines and one line with 60 units. # Example : -# Input: +# Input: # widths = [4,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10] # S = "bbbcccdddaaa" # Output: [2, 4] # -# Explanation: -# All letters except 'a' have the same length of 10, and +# Explanation: +# All letters except 'a' have the same length of 10, and # "bbbcccdddaa" will cover 9 * 10 + 2 * 4 = 98 units. # For the last 'a', it is written on the second line because # there is only 2 units left in the first line. diff --git a/Python/number-of-longest-increasing-subsequence.py b/Python/number-of-longest-increasing-subsequence.py index 9a45cadb1..3068056a1 100644 --- a/Python/number-of-longest-increasing-subsequence.py +++ b/Python/number-of-longest-increasing-subsequence.py @@ -14,7 +14,7 @@ # 5 subsequences' length is 1, so output 5. # Note: Length of the given array will be not exceed 2000 and the answer is guaranteed # to be fit in 32-bit signed int. - + class Solution(object): def findNumberOfLIS(self, nums): """ diff --git a/Python/number-of-subarrays-with-bounded-maximum.py b/Python/number-of-subarrays-with-bounded-maximum.py index 56aaaaffc..17ab89de5 100644 --- a/Python/number-of-subarrays-with-bounded-maximum.py +++ b/Python/number-of-subarrays-with-bounded-maximum.py @@ -8,7 +8,7 @@ # such that the value of the maximum array element in that subarray is at least L and at most R. # # Example : -# Input: +# Input: # A = [2, 1, 4, 3] # L = 2 # R = 3 diff --git a/Python/odd-even-linked-list.py b/Python/odd-even-linked-list.py index 457c48ac3..166f649f1 100644 --- a/Python/odd-even-linked-list.py +++ b/Python/odd-even-linked-list.py @@ -15,7 +15,7 @@ # # Note: # The relative order inside both the even and odd groups -# should remain as it was in the input. +# should remain as it was in the input. # The first node is considered odd, the second node even # and so on ... diff --git a/Python/one-edit-distance.py b/Python/one-edit-distance.py index 4e11cfe29..25a54f523 100644 --- a/Python/one-edit-distance.py +++ b/Python/one-edit-distance.py @@ -16,7 +16,7 @@ def isOneEditDistance(self, s, t): return self.isOneEditDistance(t, s) if n - m > 1: return False - + i, shift = 0, n - m while i < m and s[i] == t[i]: i += 1 @@ -24,9 +24,9 @@ def isOneEditDistance(self, s, t): i += 1 while i < m and s[i] == t[i + shift]: i += 1 - + return i == m - + if __name__ == "__main__": print Solution().isOneEditDistance("teacher", "acher") diff --git a/Python/ones-and-zeroes.py b/Python/ones-and-zeroes.py index 388bf960f..19a0b3f80 100644 --- a/Python/ones-and-zeroes.py +++ b/Python/ones-and-zeroes.py @@ -17,7 +17,7 @@ # Input: Array = {"10", "0001", "111001", "1", "0"}, m = 5, n = 3 # Output: 4 # -# Explanation: This are totally 4 strings can be formed +# Explanation: This are totally 4 strings can be formed # by the using of 5 0s and 3 1s, which are “10,”0001”,”1”,”0” # Example 2: # Input: Array = {"10", "0", "1"}, m = 1, n = 1 @@ -42,7 +42,7 @@ def findMaxForm(self, strs, m, n): elif c == '1': one_count += 1 - for i in reversed(xrange(zero_count, m+1)): - for j in reversed(xrange(one_count, n+1)): + for i in reversed(xrange(zero_count, m+1)): + for j in reversed(xrange(one_count, n+1)): dp[i][j] = max(dp[i][j], dp[i-zero_count][j-one_count]+1) return dp[m][n] diff --git a/Python/optimal-division.py b/Python/optimal-division.py index 0435f174f..5283e2e81 100644 --- a/Python/optimal-division.py +++ b/Python/optimal-division.py @@ -13,8 +13,8 @@ # Output: "1000/(100/10/2)" # Explanation: # 1000/(100/10/2) = 1000/((100/10)/2) = 200 -# However, the bold parenthesis in "1000/((100/10)/2)" are redundant, -# since they don't influence the operation priority. So you should return "1000/(100/10/2)". +# However, the bold parenthesis in "1000/((100/10)/2)" are redundant, +# since they don't influence the operation priority. So you should return "1000/(100/10/2)". # # Other cases: # 1000/(100/10)/2 = 50 diff --git a/Python/output-contest-matches.py b/Python/output-contest-matches.py index 1606a031e..8887fbf89 100644 --- a/Python/output-contest-matches.py +++ b/Python/output-contest-matches.py @@ -11,4 +11,4 @@ def findContestMatch(self, n): while len(matches)/2: matches = ["({},{})".format(matches[i], matches[-i-1]) for i in xrange(len(matches)/2)] return matches[0] - + diff --git a/Python/pacific-atlantic-water-flow.py b/Python/pacific-atlantic-water-flow.py index 57a0566e2..628dbe9d6 100644 --- a/Python/pacific-atlantic-water-flow.py +++ b/Python/pacific-atlantic-water-flow.py @@ -5,7 +5,7 @@ # each unit cell in a continent, the "Pacific ocean" touches the left and # top edges of the matrix and the "Atlantic ocean" touches the right and bottom edges. # -# Water can only flow in four directions (up, down, left, or right) +# Water can only flow in four directions (up, down, left, or right) # from a cell to another one with height equal or lower. # # Find the list of grid coordinates where water can flow to both the Pacific and Atlantic ocean. @@ -17,7 +17,7 @@ # # Given the following 5x5 matrix: # -# Pacific ~ ~ ~ ~ ~ +# Pacific ~ ~ ~ ~ ~ # ~ 1 2 2 3 (5) * # ~ 3 2 3 (4) (4) * # ~ 2 4 (5) 3 1 * @@ -44,13 +44,13 @@ def pacificAtlanticHelper(matrix, x, y, prev_height, prev_val, visited, res): (visited[x][y] | prev_val) == visited[x][y]: return - visited[x][y] |= prev_val + visited[x][y] |= prev_val if visited[x][y] == (PACIFIC | ATLANTIC): res.append((x, y)) - + for d in [(0, -1), (0, 1), (-1, 0), (1, 0)]: pacificAtlanticHelper(matrix, x + d[0], y + d[1], matrix[x][y], visited[x][y], visited, res) - + if not matrix: return [] diff --git a/Python/paint-house-ii.py b/Python/paint-house-ii.py index 724de0c42..18557f1c2 100644 --- a/Python/paint-house-ii.py +++ b/Python/paint-house-ii.py @@ -23,7 +23,7 @@ def minCostII(self, costs): """ if not costs: return 0 - + n = len(costs) k = len(costs[0]) min_cost = [costs[0], [0] * k] diff --git a/Python/paint-house.py b/Python/paint-house.py index 24e42a96b..d59b581a8 100644 --- a/Python/paint-house.py +++ b/Python/paint-house.py @@ -9,9 +9,9 @@ def minCost(self, costs): """ if not costs: return 0 - + min_cost = [costs[0], [0, 0, 0]] - + n = len(costs) for i in xrange(1, n): min_cost[i % 2][0] = costs[i][0] + \ @@ -33,11 +33,11 @@ def minCost(self, costs): """ if not costs: return 0 - + n = len(costs) for i in xrange(1, n): costs[i][0] += min(costs[i - 1][1], costs[i - 1][2]) costs[i][1] += min(costs[i - 1][0], costs[i - 1][2]) costs[i][2] += min(costs[i - 1][0], costs[i - 1][1]) - + return min(costs[n - 1]) diff --git a/Python/palindrome-number.py b/Python/palindrome-number.py index f7745350e..91a16269b 100644 --- a/Python/palindrome-number.py +++ b/Python/palindrome-number.py @@ -2,15 +2,15 @@ # Space: O(1) # # Determine whether an integer is a palindrome. Do this without extra space. -# +# # Some hints: # Could negative integers be palindromes? (ie, -1) -# +# # If you are thinking of converting the integer to string, note the restriction of using extra space. -# -# You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", +# +# You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", # you know that the reversed integer might overflow. How would you handle such case? -# +# # There is a more generic way of solving this problem. # @@ -20,12 +20,12 @@ def isPalindrome(self, x): if x < 0: return False copy, reverse = x, 0 - + while copy: reverse *= 10 reverse += copy % 10 copy //= 10 - + return x == reverse if __name__ == "__main__": diff --git a/Python/palindrome-partitioning-ii.py b/Python/palindrome-partitioning-ii.py index c2f3d98aa..192c6f0de 100644 --- a/Python/palindrome-partitioning-ii.py +++ b/Python/palindrome-partitioning-ii.py @@ -2,9 +2,9 @@ # Space: O(n^2) # # Given a string s, partition s such that every substring of the partition is a palindrome. -# +# # Return the minimum cuts needed for a palindrome partitioning of s. -# +# # For example, given s = "aab", # Return 1 since the palindrome partitioning ["aa","b"] could be produced using 1 cut. # @@ -15,13 +15,13 @@ class Solution: def minCut(self, s): lookup = [[False for j in xrange(len(s))] for i in xrange(len(s))] mincut = [len(s) - 1 - i for i in xrange(len(s) + 1)] - + for i in reversed(xrange(len(s))): for j in xrange(i, len(s)): if s[i] == s[j] and (j - i < 2 or lookup[i + 1][j - 1]): lookup[i][j] = True mincut[i] = min(mincut[i], mincut[j + 1] + 1) - + return mincut[0] if __name__ == "__main__": diff --git a/Python/palindrome-partitioning.py b/Python/palindrome-partitioning.py index fbba4054f..49bec0641 100644 --- a/Python/palindrome-partitioning.py +++ b/Python/palindrome-partitioning.py @@ -2,12 +2,12 @@ # Space: O(n^2) # # Given a string s, partition s such that every substring of the partition is a palindrome. -# +# # Return all possible palindrome partitioning of s. -# +# # For example, given s = "aab", # Return -# +# # [ # ["aa","b"], # ["a","a","b"] @@ -21,12 +21,12 @@ class Solution: # @return a list of lists of string def partition(self, s): n = len(s) - + is_palindrome = [[0 for j in xrange(n)] for i in xrange(n)] for i in reversed(xrange(0, n)): for j in xrange(i, n): is_palindrome[i][j] = s[i] == s[j] and ((j - i < 2 ) or is_palindrome[i + 1][j - 1]) - + sub_partition = [[] for i in xrange(n)] for i in reversed(xrange(n)): for j in xrange(i, n): @@ -36,7 +36,7 @@ def partition(self, s): sub_partition[i].append([s[i:j + 1]] + p) else: sub_partition[i].append([s[i:j + 1]]) - + return sub_partition[0] # Time: O(2^n) @@ -49,7 +49,7 @@ def partition(self, s): result = [] self.partitionRecu(result, [], s, 0) return result - + def partitionRecu(self, result, cur, s, i): if i == len(s): result.append(list(cur)) @@ -59,7 +59,7 @@ def partitionRecu(self, result, cur, s, i): cur.append(s[i: j + 1]) self.partitionRecu(result, cur, s, j + 1) cur.pop() - + def isPalindrome(self, s): for i in xrange(len(s) / 2): if s[i] != s[-(i + 1)]: diff --git a/Python/palindromic-substrings.py b/Python/palindromic-substrings.py index 7cc38189e..5f2668515 100644 --- a/Python/palindromic-substrings.py +++ b/Python/palindromic-substrings.py @@ -25,7 +25,7 @@ def countSubstrings(self, s): """ def manacher(s): s = '^#' + '#'.join(s) + '#$' - P = [0] * len(s) + P = [0] * len(s) C, R = 0, 0 for i in xrange(1, len(s) - 1): i_mirror = 2*C-i @@ -34,6 +34,6 @@ def manacher(s): while s[i+1+P[i]] == s[i-1-P[i]]: P[i] += 1 if i+P[i] > R: - C, R = i, i+P[i] + C, R = i, i+P[i] return P return sum((max_len+1)/2 for max_len in manacher(s)) diff --git a/Python/parse-lisp-expression.py b/Python/parse-lisp-expression.py index e0ecf7b80..24ec81409 100644 --- a/Python/parse-lisp-expression.py +++ b/Python/parse-lisp-expression.py @@ -61,7 +61,7 @@ # Explanation: Even though (let x 4 x) has a deeper scope, it is outside the context # of the final x in the add-expression. That final x will equal 2. # -# Input: (let a1 3 b2 (add a1 1) b2) +# Input: (let a1 3 b2 (add a1 1) b2) # Output 4 # Explanation: Variable names can contain digits after the first character. # diff --git a/Python/partition-list.py b/Python/partition-list.py index ec5b0a3c1..076f277b8 100644 --- a/Python/partition-list.py +++ b/Python/partition-list.py @@ -2,9 +2,9 @@ # Space: O(1) # # Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. -# +# # You should preserve the original relative order of the nodes in each of the two partitions. -# +# # For example, # Given 1->4->3->2->5->2 and x = 3, # return 1->2->2->4->3->5. @@ -15,7 +15,7 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) @@ -27,7 +27,7 @@ class Solution: def partition(self, head, x): dummySmaller, dummyGreater = ListNode(-1), ListNode(-1) smaller, greater = dummySmaller, dummyGreater - + while head: if head.val < x: smaller.next = head @@ -36,10 +36,10 @@ def partition(self, head, x): greater.next = head greater = greater.next head = head.next - + smaller.next = dummyGreater.next greater.next = None - + return dummySmaller.next if __name__ == "__main__": @@ -51,4 +51,3 @@ def partition(self, head, x): head.next.next.next.next.next = ListNode(2) print Solution().partition(head, 3) - \ No newline at end of file diff --git a/Python/partition-to-k-equal-sum-subsets.py b/Python/partition-to-k-equal-sum-subsets.py index 60ed1b5ce..6dd49fab1 100644 --- a/Python/partition-to-k-equal-sum-subsets.py +++ b/Python/partition-to-k-equal-sum-subsets.py @@ -28,15 +28,15 @@ def dfs(nums, target, used, todo, lookup): for i, num in enumerate(nums) \ if ((used>>i) & 1) == 0 and num <= targ) return lookup[used] - + total = sum(nums) if total%k or max(nums) > total//k: return False lookup = [None] * (1 << len(nums)) lookup[-1] = True return dfs(nums, total//k, 0, total, lookup) - - + + # Time: O(k^(n-k) * k!) # Space: O(n) # DFS solution with pruning. diff --git a/Python/pascals-triangle-ii.py b/Python/pascals-triangle-ii.py index 9f7cfd63d..d5aff1748 100644 --- a/Python/pascals-triangle-ii.py +++ b/Python/pascals-triangle-ii.py @@ -2,10 +2,10 @@ # Space: O(1) # Given an index k, return the kth row of the Pascal's triangle. -# +# # For example, given k = 3, # Return [1,3,3,1]. -# +# # Note: # Could you optimize your algorithm to use only O(k) extra space? # diff --git a/Python/pascals-triangle.py b/Python/pascals-triangle.py index 19b03a607..14194cc75 100644 --- a/Python/pascals-triangle.py +++ b/Python/pascals-triangle.py @@ -2,10 +2,10 @@ # Space: O(1) # # Given numRows, generate the first numRows of Pascal's triangle. -# +# # For example, given numRows = 5, # Return -# +# # [ # [1], # [1,1], diff --git a/Python/patching-array.py b/Python/patching-array.py index ce91f1f6e..8b9a681af 100644 --- a/Python/patching-array.py +++ b/Python/patching-array.py @@ -11,9 +11,9 @@ # nums = [1, 3], n = 6 # Return 1. # -# Combinations of nums are [1], [3], [1,3], which form +# Combinations of nums are [1], [3], [1,3], which form # possible sums of: 1, 3, 4. -# Now if we add/patch 2 to nums, the combinations are: +# Now if we add/patch 2 to nums, the combinations are: # [1], [2], [3], [1,3], [2,3], [1,2,3]. # Possible sums are 1, 2, 3, 4, 5, 6, which now covers # the range [1, 6]. diff --git a/Python/path-sum-ii.py b/Python/path-sum-ii.py index df35acde8..03a2fbf3a 100644 --- a/Python/path-sum-ii.py +++ b/Python/path-sum-ii.py @@ -2,7 +2,7 @@ # Space: O(h), h is height of binary tree # # Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum. -# +# # For example: # Given the below binary tree and sum = 22, # 5 @@ -33,21 +33,21 @@ class Solution: def pathSum(self, root, sum): return self.pathSumRecu([], [], root, sum) - + def pathSumRecu(self, result, cur, root, sum): if root is None: return result - + if root.left is None and root.right is None and root.val == sum: result.append(cur + [root.val]) return result - + cur.append(root.val) self.pathSumRecu(result, cur, root.left, sum - root.val) self.pathSumRecu(result, cur,root.right, sum - root.val) cur.pop() return result - + if __name__ == "__main__": root = TreeNode(5) diff --git a/Python/path-sum.py b/Python/path-sum.py index b24c951a1..70040766d 100644 --- a/Python/path-sum.py +++ b/Python/path-sum.py @@ -1,9 +1,9 @@ # Time: O(n) # Space: O(h), h is height of binary tree # -# Given a binary tree and a sum, determine if the tree has a root-to-leaf path +# Given a binary tree and a sum, determine if the tree has a root-to-leaf path # such that adding up all the values along the path equals the given sum. -# +# # For example: # Given the below binary tree and sum = 22, # 5 @@ -22,7 +22,7 @@ def __init__(self, x): self.val = x self.left = None self.right = None - + class Solution: # @param root, a tree node # @param sum, an integer @@ -30,12 +30,12 @@ class Solution: def hasPathSum(self, root, sum): if root is None: return False - + if root.left is None and root.right is None and root.val == sum: return True - + return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val) - + if __name__ == "__main__": root = TreeNode(5) root.left = TreeNode(4) diff --git a/Python/peeking-iterator.py b/Python/peeking-iterator.py index f21db1fa3..f59311e0d 100644 --- a/Python/peeking-iterator.py +++ b/Python/peeking-iterator.py @@ -48,7 +48,7 @@ def __init__(self, iterator): self.val_ = None self.has_next_ = iterator.hasNext() self.has_peeked_ = False - + def peek(self): """ diff --git a/Python/perfect-number.py b/Python/perfect-number.py index a3fefe5b5..d8d9f30cc 100644 --- a/Python/perfect-number.py +++ b/Python/perfect-number.py @@ -4,7 +4,7 @@ # We define the Perfect Number is a positive integer that is equal # to the sum of all its positive divisors except itself. # -# Now, given an integer n, write a function that returns true +# Now, given an integer n, write a function that returns true # when it is a perfect number and false when it is not. # Example: # Input: 28 @@ -20,7 +20,7 @@ def checkPerfectNumber(self, num): """ if num <= 0: return False - + sqrt_num = int(num ** 0.5) total = sum(i+num//i for i in xrange(1, sqrt_num+1) if num%i == 0) if sqrt_num ** 2 == num: diff --git a/Python/perfect-rectangle.py b/Python/perfect-rectangle.py index 7a3822f41..3cd7e513b 100644 --- a/Python/perfect-rectangle.py +++ b/Python/perfect-rectangle.py @@ -1,11 +1,11 @@ # Time: O(n) # Space: O(n) -# Given N axis-aligned rectangles where N > 0, +# Given N axis-aligned rectangles where N > 0, # determine if they all together form an exact cover of a rectangular region. # -# Each rectangle is represented as a bottom-left point and a top-right point. -# For example, a unit square is represented as [1,1,2,2]. +# Each rectangle is represented as a bottom-left point and a top-right point. +# For example, a unit square is represented as [1,1,2,2]. # (coordinate of bottom-left point is (1, 1) and top-right point is (2, 2)). # # Example 1: diff --git a/Python/perfect-squares.py b/Python/perfect-squares.py index 2cee23f58..2201bc73b 100644 --- a/Python/perfect-squares.py +++ b/Python/perfect-squares.py @@ -1,7 +1,7 @@ # Time: O(n * sqrt(n)) # Space: O(n) # -# Given a positive integer n, find the least number of perfect +# Given a positive integer n, find the least number of perfect # square numbers (for example, 1, 4, 9, 16, ...) which sum to n. # # For example, given n = 12, return 3 because 12 = 4 + 4 + 4; diff --git a/Python/permutation-sequence.py b/Python/permutation-sequence.py index cc8b59603..9382f662c 100644 --- a/Python/permutation-sequence.py +++ b/Python/permutation-sequence.py @@ -2,10 +2,10 @@ # Space: O(n) # The set [1,2,3,...,n] contains a total of n! unique permutations. -# +# # By listing and labeling all of the permutations in order, # We get the following sequence (ie, for n = 3): -# +# # "123" # "132" # "213" @@ -13,7 +13,7 @@ # "312" # "321" # Given n and k, return the kth permutation sequence. -# +# # Note: Given n will be between 1 and 9 inclusive. import math @@ -37,6 +37,6 @@ def getPermutation(self, n, k): fact /= i return seq - + if __name__ == "__main__": print Solution().getPermutation(3, 2) diff --git a/Python/permutations-ii.py b/Python/permutations-ii.py index 2b98e265c..17a03fd77 100644 --- a/Python/permutations-ii.py +++ b/Python/permutations-ii.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given a collection of numbers that might contain duplicates, return all possible unique permutations. -# +# # For example, # [1,1,2] have the following unique permutations: # [1,1,2], [1,2,1], and [2,1,1]. @@ -19,7 +19,7 @@ def permuteUnique(self, nums): used = [False] * len(nums) self.permuteUniqueRecu(result, used, [], nums) return result - + def permuteUniqueRecu(self, result, used, cur, nums): if len(cur) == len(nums): result.append(cur + []) @@ -32,13 +32,13 @@ def permuteUniqueRecu(self, result, used, cur, nums): self.permuteUniqueRecu(result, used, cur, nums) cur.pop() used[i] = False - + class Solution2: # @param num, a list of integer # @return a list of lists of integers def permuteUnique(self, nums): solutions = [[]] - + for num in nums: next = [] for solution in solutions: @@ -46,9 +46,9 @@ def permuteUnique(self, nums): candidate = solution[:i] + [num] + solution[i:] if candidate not in next: next.append(candidate) - - solutions = next - + + solutions = next + return solutions if __name__ == "__main__": diff --git a/Python/permutations.py b/Python/permutations.py index c615f05aa..0a56829ca 100644 --- a/Python/permutations.py +++ b/Python/permutations.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given a collection of numbers, return all possible permutations. -# +# # For example, # [1,2,3] have the following permutations: # [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], and [3,2,1]. @@ -17,7 +17,7 @@ def permute(self, num): used = [False] * len(num) self.permuteRecu(result, used, [], num) return result - + def permuteRecu(self, result, used, cur, num): if len(cur) == len(num): result.append(cur[:]) diff --git a/Python/populating-next-right-pointers-in-each-node-ii.py b/Python/populating-next-right-pointers-in-each-node-ii.py index e4f206e2f..2029f3293 100644 --- a/Python/populating-next-right-pointers-in-each-node-ii.py +++ b/Python/populating-next-right-pointers-in-each-node-ii.py @@ -2,11 +2,11 @@ # Space: O(1) # # Follow up for problem "Populating Next Right Pointers in Each Node". -# +# # What if the given tree could be any binary tree? Would your previous solution still work? -# +# # Note: -# +# # You may only use constant extra space. # For example, # Given the following binary tree, @@ -30,7 +30,7 @@ def __init__(self, x): self.left = None self.right = None self.next = None - + def __repr__(self): if self is None: return "Nil" @@ -50,21 +50,21 @@ def connect(self, root): next_head = cur.left elif cur.right: next_head = cur.right - + if cur.left: if prev: prev.next = cur.left prev = cur.left - + if cur.right: if prev: prev.next = cur.right prev = cur.right - + cur = cur.next head = next_head - - + + if __name__ == "__main__": root, root.left, root.right = TreeNode(1), TreeNode(2), TreeNode(3) @@ -73,4 +73,3 @@ def connect(self, root): print root print root.left print root.left.left - \ No newline at end of file diff --git a/Python/populating-next-right-pointers-in-each-node.py b/Python/populating-next-right-pointers-in-each-node.py index 79b50a7d8..54d7a63fc 100644 --- a/Python/populating-next-right-pointers-in-each-node.py +++ b/Python/populating-next-right-pointers-in-each-node.py @@ -2,18 +2,18 @@ # Space: O(1) # # Given a binary tree -# +# # struct TreeLinkNode { # TreeLinkNode *left; # TreeLinkNode *right; # TreeLinkNode *next; # } # Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL. -# +# # Initially, all next pointers are set to NULL. -# +# # Note: -# +# # You may only use constant extra space. # You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children). # For example, @@ -37,7 +37,7 @@ def __init__(self, x): self.left = None self.right = None self.next = None - + def __repr__(self): if self is None: return "Nil" @@ -81,4 +81,4 @@ def connect(self, root): print root print root.left print root.left.left - + diff --git a/Python/pour-water.py b/Python/pour-water.py index 7d331cac1..2b13679e8 100644 --- a/Python/pour-water.py +++ b/Python/pour-water.py @@ -33,7 +33,7 @@ # # w # # ## # ### # ######### -# 0123456 +# 0123456 # # When moving left or right, the water can only move to the same level or a lower level. # (By level, we mean the total height of the terrain plus any water in that column.) @@ -44,7 +44,7 @@ # # # # ## w# ### # ######### -# 0123456 +# 0123456 # # Since moving left will not make it fall, it stays in place. The next droplet falls: # @@ -52,7 +52,7 @@ # # w # # ## w# ### # ######### -# 0123456 +# 0123456 # # Since the new droplet moving left will eventually make it fall, it moves left. # Notice that the droplet still preferred to move left, @@ -62,13 +62,13 @@ # # w # # ## w# ### # ######### -# 0123456 +# 0123456 # # # # # # # # ##ww# ### # ######### -# 0123456 +# 0123456 # # After those steps, the third droplet falls. # Since moving left would not eventually make it fall, it tries to move right. @@ -78,13 +78,13 @@ # # w # # ##ww# ### # ######### -# 0123456 +# 0123456 # # # # # # # # ##ww#w### # ######### -# 0123456 +# 0123456 # # Finally, the fourth droplet falls. # Since moving left would not eventually make it fall, it tries to move right. @@ -94,14 +94,14 @@ # # w # # ##ww#w### # ######### -# 0123456 +# 0123456 # # The final answer is [2,2,2,3,2,2,2]: # -# # -# ####### -# ####### -# 0123456 +# # +# ####### +# ####### +# 0123456 # # Example 2: # Input: heights = [1,2,3,4], V = 2, K = 2 diff --git a/Python/power-of-three.py b/Python/power-of-three.py index 02f90144b..e81d1873e 100644 --- a/Python/power-of-three.py +++ b/Python/power-of-three.py @@ -21,7 +21,7 @@ def isPowerOfThree(self, n): """ return n > 0 and self.__max_pow3 % n == 0 - + class Solution2(object): def isPowerOfThree(self, n): return n > 0 and (math.log10(n)/math.log10(3)).is_integer() diff --git a/Python/predict-the-winner.py b/Python/predict-the-winner.py index b64a44cf3..6c871ef7a 100644 --- a/Python/predict-the-winner.py +++ b/Python/predict-the-winner.py @@ -13,10 +13,10 @@ # Example 1: # Input: [1, 5, 2] # Output: False -# Explanation: Initially, player 1 can choose between 1 and 2. +# Explanation: Initially, player 1 can choose between 1 and 2. # If he chooses 2 (or 1), then player 2 can choose from 1 (or 2) and 5. -# If player 2 chooses 5, then player 1 will be left with 1 (or 2). -# So, final score of player 1 is 1 + 2 = 3, and player 2 is 5. +# If player 2 chooses 5, then player 1 will be left with 1 (or 2). +# So, final score of player 1 is 1 + 2 = 3, and player 2 is 5. # Hence, player 1 will never be the winner and you need to return False. # Example 2: # Input: [1, 5, 233, 7] @@ -45,4 +45,4 @@ def PredictTheWinner(self, nums): dp[j] = max(nums[i] - dp[j], nums[j] - dp[j - 1]) return dp[-1] >= 0 - + diff --git a/Python/prime-number-of-set-bits-in-binary-representation.py b/Python/prime-number-of-set-bits-in-binary-representation.py index ce38e8afd..f843f9714 100644 --- a/Python/prime-number-of-set-bits-in-binary-representation.py +++ b/Python/prime-number-of-set-bits-in-binary-representation.py @@ -14,7 +14,7 @@ def bitCount(n): n &= n-1 result += 1 return result - + primes = {2, 3, 5, 7, 11, 13, 17, 19} return sum(bitCount(i) in primes for i in xrange(L, R+1)) diff --git a/Python/print-binary-tree.py b/Python/print-binary-tree.py index 33b00ff1d..9d7fd5460 100644 --- a/Python/print-binary-tree.py +++ b/Python/print-binary-tree.py @@ -38,10 +38,10 @@ # 1 # / \ # 2 5 -# / -# 3 -# / -# 4 +# / +# 3 +# / +# 4 # Output: # # [["", "", "", "", "", "", "", "1", "", "", "", "", "", "", ""] @@ -49,7 +49,7 @@ # ["", "3", "", "", "", "", "", "", "", "", "", "", "", "", ""] # ["4", "", "", "", "", "", "", "", "", "", "", "", "", "", ""]] # Note: The height of binary tree is in the range of [1, 10]. -# +# # Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): @@ -72,7 +72,7 @@ def getHeight(root): if not root: return 0 return max(getHeight(root.left), getHeight(root.right)) + 1 - + def preorderTraversal(root, level, left, right, result): if not root: return diff --git a/Python/product-of-array-except-self.py b/Python/product-of-array-except-self.py index f9646d83a..02b3e1a83 100644 --- a/Python/product-of-array-except-self.py +++ b/Python/product-of-array-except-self.py @@ -22,11 +22,11 @@ class Solution: def productExceptSelf(self, nums): if not nums: return [] - + left_product = [1 for _ in xrange(len(nums))] for i in xrange(1, len(nums)): left_product[i] = left_product[i - 1] * nums[i - 1] - + right_product = 1 for i in xrange(len(nums) - 2, -1, -1): right_product *= nums[i + 1] diff --git a/Python/pyramid-transition-matrix.py b/Python/pyramid-transition-matrix.py index 413a4df90..6c3073b3d 100644 --- a/Python/pyramid-transition-matrix.py +++ b/Python/pyramid-transition-matrix.py @@ -1,6 +1,6 @@ -# Time: O((a^(b+1)-a)/(a-1)) = O(a^b) , a is the size of allowed, +# Time: O((a^(b+1)-a)/(a-1)) = O(a^b) , a is the size of allowed, # b is the length of bottom -# Space: O((a^(b+1)-a)/(a-1)) = O(a^b) +# Space: O((a^(b+1)-a)/(a-1)) = O(a^b) # We are stacking blocks to form a pyramid. Each block has a color which is a one letter string, like `'Z'`. # @@ -55,7 +55,7 @@ def dfs(bottom, edges, new_bottom, idx, lookup): if dfs(bottom, edges, new_bottom, idx+1, lookup): return True return False - + if len(bottom) == 1: return True if bottom in lookup: @@ -66,7 +66,7 @@ def dfs(bottom, edges, new_bottom, idx, lookup): return False new_bottom = ['A']*(len(bottom)-1) return dfs(bottom, edges, new_bottom, 0, lookup) - + edges = [[[] for _ in xrange(7)] for _ in xrange(7)] for s in allowed: edges[ord(s[0])-ord('A')][ord(s[1])-ord('A')].append(ord(s[2])-ord('A')) diff --git a/Python/queue-reconstruction-by-height.py b/Python/queue-reconstruction-by-height.py index 7863a79ff..afd5f76cd 100644 --- a/Python/queue-reconstruction-by-height.py +++ b/Python/queue-reconstruction-by-height.py @@ -1,8 +1,8 @@ # Time: O(n * sqrt(n)) # Space: O(n) -# Suppose you have a random list of people standing in a queue. -# Each person is described by a pair of integers (h, k), +# Suppose you have a random list of people standing in a queue. +# Each person is described by a pair of integers (h, k), # where h is the height of the person and k is the number of people # in front of this person who have a height greater than or equal to h. # Write an algorithm to reconstruct the queue. @@ -29,7 +29,7 @@ def reconstructQueue(self, people): blocks = [[]] for p in people: index = p[1] - + for i, block in enumerate(blocks): if index <= len(block): break @@ -39,7 +39,7 @@ def reconstructQueue(self, people): if len(block) * len(block) > len(people): blocks.insert(i+1, block[len(block)/2:]) del block[len(block)/2:] - + return [p for block in blocks for p in block] diff --git a/Python/random-pick-index.py b/Python/random-pick-index.py index f8034c6da..b812c932e 100644 --- a/Python/random-pick-index.py +++ b/Python/random-pick-index.py @@ -27,7 +27,7 @@ class Solution(object): def __init__(self, nums): """ - + :type nums: List[int] :type numsSize: int """ diff --git a/Python/range-addition-ii.py b/Python/range-addition-ii.py index e63e660d6..641bff3db 100644 --- a/Python/range-addition-ii.py +++ b/Python/range-addition-ii.py @@ -11,22 +11,22 @@ # in the matrix after performing all the operations. # # Example 1: -# Input: +# Input: # m = 3, n = 3 # operations = [[2,2],[3,3]] # Output: 4 -# Explanation: -# Initially, M = +# Explanation: +# Initially, M = # [[0, 0, 0], # [0, 0, 0], # [0, 0, 0]] # -# After performing [2,2], M = +# After performing [2,2], M = # [[1, 1, 0], # [1, 1, 0], # [0, 0, 0]] # -# After performing [3,3], M = +# After performing [3,3], M = # [[2, 2, 1], # [2, 2, 1], # [1, 1, 1]] diff --git a/Python/range-sum-query-2d-immutable.py b/Python/range-sum-query-2d-immutable.py index 786f74939..91e444fd6 100644 --- a/Python/range-sum-query-2d-immutable.py +++ b/Python/range-sum-query-2d-immutable.py @@ -57,7 +57,7 @@ def sumRegion(self, row1, col1, row2, col2): """ return self.__sums[row2+1][col2+1] - self.__sums[row2+1][col1] - \ self.__sums[row1][col2+1] + self.__sums[row1][col1] - + # Your NumMatrix object will be instantiated and called as such: # numMatrix = NumMatrix(matrix) diff --git a/Python/range-sum-query-immutable.py b/Python/range-sum-query-immutable.py index a5c6d7775..b3b503b53 100644 --- a/Python/range-sum-query-immutable.py +++ b/Python/range-sum-query-immutable.py @@ -22,18 +22,18 @@ def __init__(self, nums): :type nums: List[int] """ self.accu = [0] - for num in nums: + for num in nums: self.accu.append(self.accu[-1] + num), def sumRange(self, i, j): """ sum of elements nums[i..j], inclusive. - :type i: int + :type i: int :type j: int - :rtype: int + :rtype: int """ return self.accu[j + 1] - self.accu[i] - + # Your NumArray object will be instantiated and called as such: # numArray = NumArray(nums) diff --git a/Python/range-sum-query-mutable.py b/Python/range-sum-query-mutable.py index d9b7db624..5e8ed8136 100644 --- a/Python/range-sum-query-mutable.py +++ b/Python/range-sum-query-mutable.py @@ -3,7 +3,7 @@ # query: O(logn) # Space: O(n) -# Given an integer array nums, find the sum of +# Given an integer array nums, find the sum of # the elements between indices i and j (i <= j), inclusive. # # The update(i, val) function modifies nums by @@ -46,7 +46,7 @@ def update(self, i, val): if val - self.__nums[i]: self.__add(i, val - self.__nums[i]) self.__nums[i] = val - + def sumRange(self, i, j): """ sum of elements nums[i..j], inclusive. @@ -87,21 +87,21 @@ def __init__(self, nums): def buildHelper(nums, start, end): if start > end: return None - + # The root's start and end is given by build method. root = self._SegmentTreeNode(start, end, 0) - + # If start equals to end, there will be no children for this node. if start == end: root.sum = nums[start] return root - + # Left child: start=nums.left, end=(nums.left + nums.right) / 2. root.left = buildHelper(nums, start, (start + end) / 2) - + # Right child: start=(nums.left + nums.right) / 2 + 1, end=nums.right. root.right = buildHelper(nums, (start + end) / 2 + 1, end) - + # Update sum. root.sum = (root.left.sum if root.left else 0) + \ (root.right.sum if root.right else 0) @@ -119,22 +119,22 @@ def updateHelper(root, i, val): # Out of range. if not root or root.start > i or root.end < i: return - + # Change the node's value with [i] to the new given value. if root.start == i and root.end == i: root.sum = val return - + updateHelper(root.left, i, val) updateHelper(root.right, i, val) - + # Update sum. root.sum = (root.left.sum if root.left else 0) + \ (root.right.sum if root.right else 0) if self.__nums[i] != val: self.__nums[i] = val updateHelper(self.__root, i, val) - + def sumRange(self, i, j): """ sum of elements nums[i..j], inclusive. diff --git a/Python/ransom-note.py b/Python/ransom-note.py index 917334555..0d175fdd4 100644 --- a/Python/ransom-note.py +++ b/Python/ransom-note.py @@ -4,7 +4,7 @@ # Given an arbitrary ransom note string and another string containing letters # from all the magazines, write a function that will return true if # the ransom note can be constructed from the magazines ; -# otherwise, it will return false. +# otherwise, it will return false. # # Each letter in the magazine string can only be used once in your ransom note. # diff --git a/Python/read-n-characters-given-read4-ii-call-multiple-times.py b/Python/read-n-characters-given-read4-ii-call-multiple-times.py index 2645c5fd0..70dccb689 100644 --- a/Python/read-n-characters-given-read4-ii-call-multiple-times.py +++ b/Python/read-n-characters-given-read4-ii-call-multiple-times.py @@ -2,11 +2,11 @@ # Space: O(1) # # The API: int read4(char *buf) reads 4 characters at a time from a file. -# +# # The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file. -# +# # By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file. -# +# # Note: # The read function may be called multiple times. # @@ -20,13 +20,13 @@ def read4(buf): while i < len(file_content) and i < 4: buf[i] = file_content[i] i += 1 - + if len(file_content) > 4: file_content = file_content[4:] else: file_content = "" return i - + # The read4 API is already defined for you. # @param buf, a list of characters # @return an integer @@ -37,7 +37,7 @@ def __init__(self): self.__buf4 = [''] * 4 self.__i4 = 0 self.__n4 = 0 - + def read(self, buf, n): """ :type buf: Destination buffer (List[str]) @@ -56,7 +56,7 @@ def read(self, buf, n): self.__i4 = 0 else: # Buffer has been empty. break - + return i if __name__ == "__main__": @@ -65,4 +65,4 @@ def read(self, buf, n): buf = ['' for _ in xrange(100)] file_content = "ab" print buf[:sol.read(buf, 1)] - print buf[:sol.read(buf, 2)] + print buf[:sol.read(buf, 2)] diff --git a/Python/read-n-characters-given-read4.py b/Python/read-n-characters-given-read4.py index 1581cca20..2c754e9f7 100644 --- a/Python/read-n-characters-given-read4.py +++ b/Python/read-n-characters-given-read4.py @@ -2,11 +2,11 @@ # Space: O(1) # # The API: int read4(char *buf) reads 4 characters at a time from a file. -# +# # The return value is the actual number of characters read. For example, it returns 3 if there is only 3 characters left in the file. -# +# # By using the read4 API, implement the function int read(char *buf, int n) that reads n characters from the file. -# +# # Note: # The read function will only be called once for each test case. # @@ -20,13 +20,13 @@ def read4(buf): while i < len(file_content) and i < 4: buf[i] = file_content[i] i += 1 - + if len(file_content) > 4: file_content = file_content[4:] else: file_content = "" return i - + class Solution(object): def read(self, buf, n): """ @@ -49,6 +49,6 @@ def read(self, buf, n): global file_content buf = ['' for _ in xrange(100)] file_content = "a" - print buf[:Solution().read(buf, 9)] + print buf[:Solution().read(buf, 9)] file_content = "abcdefghijklmnop" print buf[:Solution().read(buf, 9)] diff --git a/Python/reconstruct-original-digits-from-english.py b/Python/reconstruct-original-digits-from-english.py index ff7797bde..eb67abeab 100644 --- a/Python/reconstruct-original-digits-from-english.py +++ b/Python/reconstruct-original-digits-from-english.py @@ -30,10 +30,10 @@ def originalDigits(self, s): cnts = [Counter(_) for _ in ["zero", "one", "two", "three", \ "four", "five", "six", "seven", \ "eight", "nine"]] - + # The order for greedy method. order = [0, 2, 4, 6, 8, 1, 3, 5, 7, 9] - + # The unique char in the order. unique_chars = ['z', 'o', 'w', 't', 'u', \ 'f', 'x', 's', 'g', 'n'] diff --git a/Python/recover-binary-search-tree.py b/Python/recover-binary-search-tree.py index 0cc9ddda1..073967d9e 100644 --- a/Python/recover-binary-search-tree.py +++ b/Python/recover-binary-search-tree.py @@ -2,9 +2,9 @@ # Space: O(1) # # Two elements of a binary search tree (BST) are swapped by mistake. -# +# # Recover the tree without changing its structure. -# +# # Note: # A solution using O(n) space is pretty straight forward. Could you devise a constant space solution? # @@ -14,8 +14,8 @@ class TreeNode: def __init__(self, x): self.val = x self.left = None - self.right = None - + self.right = None + def __repr__(self): if self: serial = [] @@ -23,21 +23,21 @@ def __repr__(self): while queue: cur = queue[0] - + if cur: serial.append(cur.val) queue.append(cur.left) queue.append(cur.right) else: serial.append("#") - + queue = queue[1:] - + while serial[-1] == "#": serial.pop() - + return repr(serial) - + else: return None @@ -46,13 +46,13 @@ class Solution: # @return a tree node def recoverTree(self, root): return self.MorrisTraversal(root) - + def MorrisTraversal(self, root): if root is None: return broken = [None, None] pre, cur = None, root - + while cur: if cur.left is None: self.detectBroken(broken, pre, cur) @@ -62,7 +62,7 @@ def MorrisTraversal(self, root): node = cur.left while node.right and node.right != cur: node = node.right - + if node.right is None: node.right =cur cur = cur.left @@ -71,20 +71,19 @@ def MorrisTraversal(self, root): node.right = None pre = cur cur = cur.right - + broken[0].val, broken[1].val = broken[1].val, broken[0].val - + return root - + def detectBroken(self, broken, pre, cur): if pre and pre.val > cur.val: if broken[0] is None: broken[0] = pre broken[1] = cur - + if __name__ == "__main__": root = TreeNode(0) root.left = TreeNode(1) print root print Solution().recoverTree(root) - \ No newline at end of file diff --git a/Python/redundant-connection-ii.py b/Python/redundant-connection-ii.py index 39f86635f..79e613c68 100644 --- a/Python/redundant-connection-ii.py +++ b/Python/redundant-connection-ii.py @@ -56,7 +56,7 @@ def union_set(self, x, y): self.count -= 1 return True - + class Solution(object): def findRedundantDirectedConnection(self, edges): """ diff --git a/Python/redundant-connection.py b/Python/redundant-connection.py index c26efe3dd..c490e4743 100644 --- a/Python/redundant-connection.py +++ b/Python/redundant-connection.py @@ -59,4 +59,4 @@ def findRedundantConnection(self, edges): if not union_find.union_set(*edge): return edge return [] - + diff --git a/Python/regular-expression-matching.py b/Python/regular-expression-matching.py index 3101c52c7..ef240b84a 100644 --- a/Python/regular-expression-matching.py +++ b/Python/regular-expression-matching.py @@ -2,15 +2,15 @@ # Space: O(n) # # Implement regular expression matching with support for '.' and '*'. -# +# # '.' Matches any single character. # '*' Matches zero or more of the preceding element. -# +# # The matching should cover the entire input string (not partial). -# +# # The function prototype should be: # bool isMatch(const char *s, const char *p) -# +# # Some examples: # isMatch("aa","a") -> false # isMatch("aa","aa") -> true @@ -27,12 +27,12 @@ class Solution: def isMatch(self, s, p): k = 3 result = [[False for j in xrange(len(p) + 1)] for i in xrange(k)] - + result[0][0] = True for i in xrange(2, len(p) + 1): if p[i-1] == '*': result[0][i] = result[0][i-2] - + for i in xrange(1,len(s) + 1): if i > 1: result[0][0] = False @@ -41,7 +41,7 @@ def isMatch(self, s, p): result[i % k][j] = result[(i-1) % k][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.') else: result[i % k][j] = result[i % k][j-2] or (result[(i-1) % k][j] and (s[i-1] == p[j-2] or p[j-2] == '.')) - + return result[len(s) % k][len(p)] # dp @@ -51,19 +51,19 @@ class Solution2: # @return a boolean def isMatch(self, s, p): result = [[False for j in xrange(len(p) + 1)] for i in xrange(len(s) + 1)] - + result[0][0] = True for i in xrange(2, len(p) + 1): if p[i-1] == '*': result[0][i] = result[0][i-2] - + for i in xrange(1,len(s) + 1): for j in xrange(1, len(p) + 1): if p[j-1] != '*': result[i][j] = result[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '.') else: result[i][j] = result[i][j-2] or (result[i-1][j] and (s[i-1] == p[j-2] or p[j-2] == '.')) - + return result[len(s)][len(p)] # iteration @@ -84,7 +84,7 @@ def isMatch(self, s, p): [last_s_ptr, last_p_ptr] = last_ptr.pop() while last_ptr and p[last_p_ptr - 2] != s[last_s_ptr] and p[last_p_ptr - 2] != '.': [last_s_ptr, last_p_ptr] = last_ptr.pop() - + if p[last_p_ptr - 2] == s[last_s_ptr] or p[last_p_ptr - 2] == '.': last_s_ptr += 1 s_ptr = last_s_ptr @@ -94,19 +94,19 @@ def isMatch(self, s, p): return False else: return False - + while p_ptr < len(p) - 1 and p[p_ptr] == '.' and p[p_ptr + 1] == '*': p_ptr += 2 - + return p_ptr == len(p) - + # recursive class Solution4: # @return a boolean def isMatch(self, s, p): if not p: return not s - + if len(p) == 1 or p[1] != '*': if len(s) > 0 and (p[0] == s[0] or p[0] == '.'): return self.isMatch(s[1:], p[1:]) diff --git a/Python/relative-ranks.py b/Python/relative-ranks.py index 3b32dd9a4..35b12842f 100644 --- a/Python/relative-ranks.py +++ b/Python/relative-ranks.py @@ -9,7 +9,7 @@ # Input: [5, 4, 3, 2, 1] # Output: ["Gold Medal", "Silver Medal", "Bronze Medal", "4", "5"] # Explanation: The first three athletes got the top three highest scores, -# so they got "Gold Medal", "Silver Medal" and "Bronze Medal". +# so they got "Gold Medal", "Silver Medal" and "Bronze Medal". # For the left two athletes, you just need to output # their relative ranks according to their scores. # Note: diff --git a/Python/remove-9.py b/Python/remove-9.py index ec7ad01e3..234f57714 100644 --- a/Python/remove-9.py +++ b/Python/remove-9.py @@ -1,6 +1,6 @@ # Time: O(logn) # Space: O(1) - + class Solution(object): def newInteger(self, n): """ diff --git a/Python/remove-boxes.py b/Python/remove-boxes.py index a34b96d6a..e1012647e 100644 --- a/Python/remove-boxes.py +++ b/Python/remove-boxes.py @@ -1,7 +1,7 @@ # Time: O(n^3) ~ O(n^4) # Space: O(n^3) -# Given several boxes with different colors represented by different positive numbers. +# Given several boxes with different colors represented by different positive numbers. # You may experience several rounds to remove boxes until there is no box left. # Each time you can choose some continuous boxes with the same color (composed of k boxes, k >= 1), # remove them and get k*k points. @@ -14,10 +14,10 @@ # Output: # 23 # Explanation: -# [1, 3, 2, 2, 2, 3, 4, 3, 1] -# ----> [1, 3, 3, 4, 3, 1] (3*3=9 points) -# ----> [1, 3, 3, 3, 1] (1*1=1 points) -# ----> [1, 1] (3*3=9 points) +# [1, 3, 2, 2, 2, 3, 4, 3, 1] +# ----> [1, 3, 3, 4, 3, 1] (3*3=9 points) +# ----> [1, 3, 3, 3, 1] (1*1=1 points) +# ----> [1, 1] (3*3=9 points) # ----> [] (2*2=4 points) # Note: The number of boxes n would not exceed 100. diff --git a/Python/remove-comments.py b/Python/remove-comments.py index 5b8707e0e..57895225a 100644 --- a/Python/remove-comments.py +++ b/Python/remove-comments.py @@ -11,15 +11,15 @@ # rest of the characters to the right of it in the same line should be ignored. # # The string /* denotes a block comment, -# which represents that all characters until the next (non-overlapping) occurrence of */ -# should be ignored. (Here, occurrences happen in reading order: line by line from left to right.) +# which represents that all characters until the next (non-overlapping) occurrence of */ +# should be ignored. (Here, occurrences happen in reading order: line by line from left to right.) # To be clear, the string /*/ does not yet end the block comment, as the ending would be overlapping the beginning. # # The first effective comment takes precedence over others: # if the string // occurs in a block comment, it is ignored. # Similarly, if the string /* occurs in a line or block comment, it is also ignored. # -# If a certain line of code is empty after removing comments, +# If a certain line of code is empty after removing comments, # you must not output that line: each string in the answer list will be non-empty. # # There will be no control characters, single quote, or double quote characters. @@ -34,18 +34,18 @@ # After removing the comments from the source code, return the source code in the same format. # # Example 1: -# Input: +# Input: # source = ["/*Test program */", "int main()", "{ ", " // variable declaration ", "int a, b, c;", "/* This is a test", " multiline ", " comment for ", " testing */", "a = b + c;", "}"] # # The line by line code is visualized as below: # /*Test program */ # int main() -# { -# // variable declaration +# { +# // variable declaration # int a, b, c; # /* This is a test -# multiline -# comment for +# multiline +# comment for # testing */ # a = b + c; # } @@ -54,19 +54,19 @@ # # The line by line code is visualized as below: # int main() -# { +# { # int a, b, c; # a = b + c; # } -# Explanation: -# The string +# Explanation: +# The string # /* -# denotes a block comment, including line 1 and lines 6-9. The string +# denotes a block comment, including line 1 and lines 6-9. The string # // # denotes line 4 as comments. # # Example 2: -# Input: +# Input: # source = ["a/*comment", "line", "more_comment*/b"] # Output: ["ab"] # Explanation: The original source string is "a/*comment\nline\nmore_comment*/b", diff --git a/Python/remove-duplicates-from-sorted-array-ii.py b/Python/remove-duplicates-from-sorted-array-ii.py index 794b192ae..104161e0a 100644 --- a/Python/remove-duplicates-from-sorted-array-ii.py +++ b/Python/remove-duplicates-from-sorted-array-ii.py @@ -3,10 +3,10 @@ # # Follow up for "Remove Duplicates": # What if duplicates are allowed at most twice? -# +# # For example, # Given sorted array A = [1,1,1,2,2,3], -# +# # Your function should return length = 5, and A is now [1,1,2,2,3]. # @@ -16,7 +16,7 @@ class Solution: def removeDuplicates(self, A): if not A: return 0 - + last, i, same = 0, 1, False while i < len(A): if A[last] != A[i] or not same: @@ -24,7 +24,7 @@ def removeDuplicates(self, A): last += 1 A[last] = A[i] i += 1 - + return last + 1 if __name__ == "__main__": diff --git a/Python/remove-duplicates-from-sorted-array.py b/Python/remove-duplicates-from-sorted-array.py index 7ea358680..b2a94cc23 100644 --- a/Python/remove-duplicates-from-sorted-array.py +++ b/Python/remove-duplicates-from-sorted-array.py @@ -2,12 +2,12 @@ # Space: O(1) # # Given a sorted array, remove the duplicates in place such that each element appear only once and return the new length. -# +# # Do not allocate extra space for another array, you must do this in place with constant memory. -# +# # For example, # Given input array A = [1,1,2], -# +# # Your function should return length = 2, and A is now [1,2]. # @@ -17,14 +17,14 @@ class Solution: def removeDuplicates(self, A): if not A: return 0 - + last, i = 0, 1 while i < len(A): if A[last] != A[i]: last += 1 A[last] = A[i] i += 1 - + return last + 1 if __name__ == "__main__": diff --git a/Python/remove-duplicates-from-sorted-list-ii.py b/Python/remove-duplicates-from-sorted-list-ii.py index 0f21f2171..9d4d364be 100644 --- a/Python/remove-duplicates-from-sorted-list-ii.py +++ b/Python/remove-duplicates-from-sorted-list-ii.py @@ -3,7 +3,7 @@ # # Given a sorted linked list, delete all nodes that have duplicate numbers, # leaving only distinct numbers from the original list. -# +# # For example, # Given 1->2->3->3->4->4->5, return 1->2->5. # Given 1->1->1->2->3, return 2->3. @@ -14,7 +14,7 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self is None: return "Nil" @@ -40,10 +40,10 @@ def deleteDuplicates(self, head): pre = cur cur = cur.next return dummy.next - + if __name__ == "__main__": head, head.next, head.next.next = ListNode(1), ListNode(2), ListNode(3) head.next.next.next, head.next.next.next.next = ListNode(3), ListNode(4) head.next.next.next.next.next, head.next.next.next.next.next.next = ListNode(4), ListNode(5) print Solution().deleteDuplicates(head) - + diff --git a/Python/remove-element.py b/Python/remove-element.py index a24b77318..edb0f237b 100644 --- a/Python/remove-element.py +++ b/Python/remove-element.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given an array and a value, remove all instances of that value in place and return the new length. -# +# # The order of elements can be changed. It doesn't matter what you leave beyond the new length. # @@ -19,6 +19,6 @@ def removeElement(self, A, elem): else: i += 1 return last + 1 - + if __name__ == "__main__": print Solution().removeElement([1, 2, 3, 4, 5, 2, 2], 2) \ No newline at end of file diff --git a/Python/remove-invalid-parentheses.py b/Python/remove-invalid-parentheses.py index 35504e3ca..348e4291e 100644 --- a/Python/remove-invalid-parentheses.py +++ b/Python/remove-invalid-parentheses.py @@ -54,7 +54,7 @@ def removeInvalidParenthesesHelper(start, left_removed, right_removed): if isValid(tmp): res.append(tmp) return - + for i in xrange(start, len(s)): if right_removed == 0 and left_removed > 0 and s[i] == '(': if i == start or s[i] != s[i - 1]: # Skip duplicated. diff --git a/Python/remove-k-digits.py b/Python/remove-k-digits.py index ec30523a3..b9b9ee749 100644 --- a/Python/remove-k-digits.py +++ b/Python/remove-k-digits.py @@ -16,7 +16,7 @@ # # Input: num = "10200", k = 1 # Output: "200" -# Explanation: Remove the leading 1 and the number is 200. +# Explanation: Remove the leading 1 and the number is 200. # Note that the output must not contain leading zeroes. # Example 3: # diff --git a/Python/remove-linked-list-elements.py b/Python/remove-linked-list-elements.py index 347370e88..316f1d6de 100644 --- a/Python/remove-linked-list-elements.py +++ b/Python/remove-linked-list-elements.py @@ -21,15 +21,15 @@ def removeElements(self, head, val): dummy = ListNode(float("-inf")) dummy.next = head prev, curr = dummy, dummy.next - + while curr: if curr.val == val: prev.next = curr.next else: prev = curr - + curr = curr.next - + return dummy.next - - + + diff --git a/Python/remove-nth-node-from-end-of-list.py b/Python/remove-nth-node-from-end-of-list.py index ea655cd08..637c8c198 100644 --- a/Python/remove-nth-node-from-end-of-list.py +++ b/Python/remove-nth-node-from-end-of-list.py @@ -2,11 +2,11 @@ # Space: O(1) # # Given a linked list, remove the nth node from the end of list and return its head. -# +# # For example, -# +# # Given linked list: 1->2->3->4->5, and n = 2. -# +# # After removing the second node from the end, the linked list becomes 1->2->3->5. # Note: # Given n will always be valid. @@ -24,22 +24,22 @@ def __repr__(self): return "Nil" else: return "{} -> {}".format(self.val, repr(self.next)) - + class Solution: # @return a ListNode def removeNthFromEnd(self, head, n): dummy = ListNode(-1) dummy.next = head slow, fast = dummy, dummy - + for i in xrange(n): fast = fast.next - + while fast.next: slow, fast = slow.next, fast.next - + slow.next = slow.next.next - + return dummy.next if __name__ == "__main__": @@ -48,5 +48,5 @@ def removeNthFromEnd(self, head, n): head.next.next = ListNode(3) head.next.next.next = ListNode(4) head.next.next.next.next = ListNode(5) - + print Solution().removeNthFromEnd(head, 2) \ No newline at end of file diff --git a/Python/reorder-list.py b/Python/reorder-list.py index a80c0fede..8d5541c83 100644 --- a/Python/reorder-list.py +++ b/Python/reorder-list.py @@ -3,9 +3,9 @@ # # Given a singly linked list L: L0->L1->...->Ln-1->Ln, # reorder it to: L0->Ln->L1->Ln-1->L2->Ln-2->... -# +# # You must do this in-place without altering the nodes' values. -# +# # For example, # Given {1,2,3,4}, reorder it to {1,4,2,3}. # @@ -15,26 +15,26 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) - + class Solution: # @param head, a ListNode # @return nothing def reorderList(self, head): if head == None or head.next == None: return head - + fast, slow, prev = head, head, None while fast != None and fast.next != None: fast, slow, prev = fast.next.next, slow.next, slow current, prev.next, prev = slow, None, None - + while current != None: current.next, prev, current = prev, current, current.next - + l1, l2 = head, prev dummy = ListNode(0) current = dummy @@ -42,7 +42,7 @@ def reorderList(self, head): while l1 != None and l2 != None: current.next, current, l1 = l1, l1, l1.next current.next, current, l2 = l2, l2, l2.next - + return dummy.next if __name__ == "__main__": diff --git a/Python/repeated-dna-sequences.py b/Python/repeated-dna-sequences.py index 901af0451..bb9373bfa 100644 --- a/Python/repeated-dna-sequences.py +++ b/Python/repeated-dna-sequences.py @@ -1,9 +1,9 @@ # Time: O(n) # Space: O(n) -# All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, +# All DNA is composed of a series of nucleotides abbreviated as A, C, G, and T, # for example: "ACGAATTCCG". When studying DNA, it is sometimes useful to identify repeated sequences within the DNA. -# +# # Write a function to find all the 10-letter-long sequences (substrings) that occur more than once in a DNA molecule. # # For example, diff --git a/Python/reshape-the-matrix.py b/Python/reshape-the-matrix.py index acbe6bdb2..6c5eba08e 100644 --- a/Python/reshape-the-matrix.py +++ b/Python/reshape-the-matrix.py @@ -15,24 +15,24 @@ # output the new reshaped matrix; Otherwise, output the original matrix. # # Example 1: -# Input: -# nums = +# Input: +# nums = # [[1,2], # [3,4]] # r = 1, c = 4 -# Output: +# Output: # [[1,2,3,4]] # Explanation: # The row-traversing of nums is [1,2,3,4]. The new reshaped matrix is a 1 * 4 matrix, # fill it row by row by using the previous list. # # Example 2: -# Input: -# nums = +# Input: +# nums = # [[1,2], # [3,4]] # r = 2, c = 4 -# Output: +# Output: # [[1,2], # [3,4]] # Explanation: diff --git a/Python/restore-ip-addresses.py b/Python/restore-ip-addresses.py index 504ce6c6c..e1e338b64 100644 --- a/Python/restore-ip-addresses.py +++ b/Python/restore-ip-addresses.py @@ -2,10 +2,10 @@ # Space: O(n * m) = O(3 * 4) # # Given a string containing only digits, restore it by returning all possible valid IP address combinations. -# +# # For example: # Given "25525511135", -# +# # return ["255.255.11.135", "255.255.111.35"]. (Order does not matter) # @@ -16,12 +16,12 @@ def restoreIpAddresses(self, s): result = [] self.restoreIpAddressesRecur(result, s, 0, "", 0) return result - + def restoreIpAddressesRecur(self, result, s, start, current, dots): # pruning to improve performance if (4 - dots) * 3 < len(s) - start or (4 - dots) > len(s) - start: return - + if start == len(s) and dots == 4: result.append(current[:-1]) else: @@ -30,11 +30,11 @@ def restoreIpAddressesRecur(self, result, s, start, current, dots): current += s[start:i + 1] + '.' self.restoreIpAddressesRecur(result, s, i + 1, current, dots + 1) current = current[:-(i - start + 2)] - + def isValid(self, s): if len(s) == 0 or (s[0] == '0' and s != "0"): return False return int(s) < 256 - + if __name__ == "__main__": print Solution().restoreIpAddresses("0000") diff --git a/Python/reverse-bits.py b/Python/reverse-bits.py index 636684e0e..baecdb400 100644 --- a/Python/reverse-bits.py +++ b/Python/reverse-bits.py @@ -29,6 +29,6 @@ def reverseBits2(self, n): else: string = string[:2] + string[2:].zfill(32)[::-1] return int(string, 2) - + if __name__ == '__main__': print Solution().reverseBits(1) diff --git a/Python/reverse-integer.py b/Python/reverse-integer.py index 6b213a576..8533bbe1c 100644 --- a/Python/reverse-integer.py +++ b/Python/reverse-integer.py @@ -2,21 +2,21 @@ # Space: O(1) # # Reverse digits of an integer. -# +# # Example1: x = 123, return 321 # Example2: x = -123, return -321 -# +# # click to show spoilers. -# +# # Have you thought about this? # Here are some good questions to ask before coding. Bonus points for you if you have already thought through this! -# +# # If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100. -# -# Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, +# +# Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, # then the reverse of 1000000003 overflows. How should you handle such cases? -# -# Throw an exception? Good, but what if throwing an exception is not an option? +# +# Throw an exception? Good, but what if throwing an exception is not an option? # You would then have to re-design the function (ie, add an extra parameter). @@ -34,7 +34,7 @@ def reverse(self, x): result = result * 10 + x % 10 x //= 10 return result if result <= 0x7fffffff else 0 # Handle overflow. - + def reverse2(self, x): """ :type x: int diff --git a/Python/reverse-linked-list-ii.py b/Python/reverse-linked-list-ii.py index 4314d1b6f..d18404785 100644 --- a/Python/reverse-linked-list-ii.py +++ b/Python/reverse-linked-list-ii.py @@ -2,12 +2,12 @@ # Space: O(1) # # Reverse a linked list from position m to n. Do it in-place and in one-pass. -# +# # For example: # Given 1->2->3->4->5->NULL, m = 2 and n = 4, -# +# # return 1->4->3->2->5->NULL. -# +# # Note: # Given m, n satisfy the following condition: # 1 <= m <= n <= length of list. @@ -18,7 +18,7 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) @@ -30,17 +30,17 @@ class Solution: def reverseBetween(self, head, m, n): diff, dummy, cur = n - m + 1, ListNode(-1), head dummy.next = head - + last_unswapped = dummy while cur and m > 1: cur, last_unswapped, m = cur.next, cur, m - 1 - + prev, first_swapped = last_unswapped, cur while cur and diff > 0: cur.next, prev, cur, diff = prev, cur, cur.next, diff - 1 - + last_unswapped.next, first_swapped.next = prev, cur - + return dummy.next if __name__ == "__main__": @@ -50,4 +50,3 @@ def reverseBetween(self, head, m, n): head.next.next.next = ListNode(4) head.next.next.next.next = ListNode(5) print Solution().reverseBetween(head, 2, 4) - \ No newline at end of file diff --git a/Python/reverse-linked-list.py b/Python/reverse-linked-list.py index 8e8441ddc..b000888f2 100644 --- a/Python/reverse-linked-list.py +++ b/Python/reverse-linked-list.py @@ -14,7 +14,7 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) @@ -31,27 +31,27 @@ def reverseList(self, head): # Time: O(n) # Space: O(n) -# Recursive solution. +# Recursive solution. class Solution2: # @param {ListNode} head # @return {ListNode} def reverseList(self, head): [begin, end] = self.reverseListRecu(head) return begin - + def reverseListRecu(self, head): if not head: return [None, None] - + [begin, end] = self.reverseListRecu(head.next) - + if end: end.next = head head.next = None return [begin, head] else: return [head, head] - + if __name__ == "__main__": head = ListNode(1) head.next = ListNode(2) diff --git a/Python/reverse-nodes-in-k-group.py b/Python/reverse-nodes-in-k-group.py index 359fce169..15f95d5fa 100644 --- a/Python/reverse-nodes-in-k-group.py +++ b/Python/reverse-nodes-in-k-group.py @@ -2,18 +2,18 @@ # Space: O(1) # # Given a linked list, reverse the nodes of a linked list k at a time and return its modified list. -# +# # If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is. -# +# # You may not alter the values in the nodes, only nodes itself may be changed. -# +# # Only constant memory is allowed. -# +# # For example, # Given this linked list: 1->2->3->4->5 -# +# # For k = 2, you should return: 2->1->4->3->5 -# +# # For k = 3, you should return: 3->2->1->4->5 # @@ -21,8 +21,8 @@ class ListNode: def __init__(self, x): self.val = x - self.next = None - + self.next = None + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) @@ -34,27 +34,27 @@ class Solution: def reverseKGroup(self, head, k): dummy = ListNode(-1) dummy.next = head - + cur, cur_dummy = head, dummy length = 0 - + while cur: next_cur = cur.next length = (length + 1) % k - + if length == 0: next_dummy = cur_dummy.next self.reverse(cur_dummy, cur.next) cur_dummy = next_dummy - + cur = next_cur - + return dummy.next - + def reverse(self, begin, end): first = begin.next cur = first.next - + while cur != end: first.next = cur.next cur.next = begin.next diff --git a/Python/reverse-vowels-of-a-string.py b/Python/reverse-vowels-of-a-string.py index f7eb88ae2..359c71f8a 100644 --- a/Python/reverse-vowels-of-a-string.py +++ b/Python/reverse-vowels-of-a-string.py @@ -1,7 +1,7 @@ # Time: O(n) # Space: O(1) -# Write a function that takes a string as input +# Write a function that takes a string as input # and reverse only the vowels of a string. # # Example 1: diff --git a/Python/reverse-words-in-a-string-ii.py b/Python/reverse-words-in-a-string-ii.py index 6656ae327..e87268406 100644 --- a/Python/reverse-words-in-a-string-ii.py +++ b/Python/reverse-words-in-a-string-ii.py @@ -3,14 +3,14 @@ # # Given an input string, reverse the string word by word. # A word is defined as a sequence of non-space characters. -# +# # The input string does not contain leading or trailing spaces # and the words are always separated by a single space. -# +# # For example, # Given s = "the sky is blue", # return "blue is sky the". -# +# # Could you do it in-place without allocating extra space? # diff --git a/Python/reverse-words-in-a-string.py b/Python/reverse-words-in-a-string.py index 4d4bddc90..e660e205c 100644 --- a/Python/reverse-words-in-a-string.py +++ b/Python/reverse-words-in-a-string.py @@ -2,13 +2,13 @@ # Space: O(n) # # Given an input string, reverse the string word by word. -# +# # For example, # Given s = "the sky is blue", # return "blue is sky the". -# +# # click to show clarification. -# +# # Clarification: # What constitutes a word? # A sequence of non-space characters constitutes a word. diff --git a/Python/roman-to-integer.py b/Python/roman-to-integer.py index 2b858fea1..a5d2d05d9 100644 --- a/Python/roman-to-integer.py +++ b/Python/roman-to-integer.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a roman numeral, convert it to an integer. -# +# # Input is guaranteed to be within the xrange from 1 to 3999. # diff --git a/Python/rotate-array.py b/Python/rotate-array.py index 1139d6a33..7e5484fca 100644 --- a/Python/rotate-array.py +++ b/Python/rotate-array.py @@ -60,7 +60,7 @@ def apply_cycle_permutation(self, k, offset, cycle_len, nums): nums[(offset + i * k) % len(nums)], tmp = tmp, nums[(offset + i * k) % len(nums)] nums[offset] = tmp - + class Solution3: """ :type nums: List[int] @@ -82,7 +82,7 @@ def rotate(self, nums, k): if start == curr: break start += 1 - + if __name__ == '__main__': nums = [1, 2, 3, 4, 5, 6, 7] diff --git a/Python/rotate-image.py b/Python/rotate-image.py index dabc2734f..648b4c715 100644 --- a/Python/rotate-image.py +++ b/Python/rotate-image.py @@ -2,9 +2,9 @@ # Space: O(1) # # You are given an n x n 2D matrix representing an image. -# +# # Rotate the image by 90 degrees (clockwise). -# +# # Follow up: # Could you do this in-place? # @@ -16,19 +16,19 @@ class Solution: # @return a list of lists of integers def rotate(self, matrix): n = len(matrix) - + # anti-diagonal mirror for i in xrange(n): for j in xrange(n - i): matrix[i][j], matrix[n-1-j][n-1-i] = matrix[n-1-j][n-1-i], matrix[i][j] - + # horizontal mirror for i in xrange(n / 2): for j in xrange(n): matrix[i][j], matrix[n-1-i][j] = matrix[n-1-i][j], matrix[i][j] - + return matrix - + # Time: O(n^2) # Space: O(n^2) class Solution2: @@ -36,7 +36,7 @@ class Solution2: # @return a list of lists of integers def rotate(self, matrix): return [list(reversed(x)) for x in zip(*matrix)] - + if __name__ == "__main__": matrix = [[1, 2, 3], [4, 5, 6], [7, 8, 9]] print Solution().rotate(matrix) diff --git a/Python/rotate-list.py b/Python/rotate-list.py index 6a536da17..59ecb567e 100644 --- a/Python/rotate-list.py +++ b/Python/rotate-list.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a list, rotate the list to the right by k places, where k is non-negative. -# +# # For example: # Given 1->2->3->4->5->NULL and k = 2, # return 4->5->1->2->3->NULL. @@ -13,11 +13,11 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) - + class Solution(object): def rotateRight(self, head, k): """ @@ -42,7 +42,7 @@ def rotateRight(self, head, k): return cur - + if __name__ == "__main__": head = ListNode(1) head.next = ListNode(2) diff --git a/Python/rotate-string.py b/Python/rotate-string.py index b988e3bf4..3a8f5151a 100644 --- a/Python/rotate-string.py +++ b/Python/rotate-string.py @@ -32,7 +32,7 @@ def check(index): if len(A) != len(B): return False - + M, p = 10**9+7, 113 p_inv = pow(p, M-2, M) @@ -98,12 +98,12 @@ def getPrefix(pattern): if not needle: return 0 return KMP(haystack, needle) - + if len(A) != len(B): return False return strStr(A*2, B) != -1 - + # Time: O(n^2) # Space: O(n) class Solution3(object): diff --git a/Python/rotated-digits.py b/Python/rotated-digits.py index 4a0f1baf4..98223aa44 100644 --- a/Python/rotated-digits.py +++ b/Python/rotated-digits.py @@ -12,7 +12,7 @@ # Example: # Input: 10 # Output: 4 -# Explanation: +# Explanation: # There are four good numbers in the range [1, 10] : 2, 5, 6, 9. # Note that 1 and 10 are not good numbers, since they remain unchanged after rotating. # @@ -40,11 +40,11 @@ def dp(A, i, is_prefix_equal, is_good, lookup): lookup) lookup[i, is_prefix_equal, is_good] = result return lookup[i, is_prefix_equal, is_good] - + lookup = {} return dp(A, 0, True, False, lookup) - - + + # Time: O(n) # Space: O(n) class Solution2(object): @@ -66,7 +66,7 @@ def rotatedDigits(self, N): if i*10+j <= N: dp[i*10+j] = DIFF return dp.count(DIFF) - + # Time: O(nlogn) = O(n), because O(logn) = O(32) by this input # Space: O(logn) = O(1) @@ -85,4 +85,4 @@ def rotatedDigits(self, N): if diff & lookup: result += 1 return result - + diff --git a/Python/russian-doll-envelopes.py b/Python/russian-doll-envelopes.py index 2f695b9c4..50685c92b 100644 --- a/Python/russian-doll-envelopes.py +++ b/Python/russian-doll-envelopes.py @@ -3,7 +3,7 @@ # You have a number of envelopes with widths and heights given # as a pair of integers (w, h). One envelope can fit into another -# if and only if both the width and height of one envelope is greater +# if and only if both the width and height of one envelope is greater # than the width and height of the other envelope. # # What is the maximum number of envelopes can you Russian doll? diff --git a/Python/same-tree.py b/Python/same-tree.py index c7e799c94..6418ffc0e 100644 --- a/Python/same-tree.py +++ b/Python/same-tree.py @@ -2,7 +2,7 @@ # Space: O(h), h is height of binary tree # # Given two binary trees, write a function to check if they are equal or not. -# +# # Two binary trees are considered equal if they are structurally identical and the nodes have the same value. # @@ -20,12 +20,12 @@ class Solution: def isSameTree(self, p, q): if p is None and q is None: return True - + if p is not None and q is not None: return p.val == q.val and self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right) - + return False - + if __name__ == "__main__": root1, root1.left, root1.right = TreeNode(1), TreeNode(2), TreeNode(3) root2, root2.left, root2.right = TreeNode(1), TreeNode(2), TreeNode(3) diff --git a/Python/scramble-string.py b/Python/scramble-string.py index 7969954b0..2627135af 100644 --- a/Python/scramble-string.py +++ b/Python/scramble-string.py @@ -2,9 +2,9 @@ # Space: O(n^3) # # Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively. -# +# # Below is one possible representation of s1 = "great": -# +# # great # / \ # gr eat @@ -13,9 +13,9 @@ # / \ # a t # To scramble the string, we may choose any non-leaf node and swap its two children. -# +# # For example, if we choose the node "gr" and swap its two children, it produces a scrambled string "rgeat". -# +# # rgeat # / \ # rg eat @@ -24,9 +24,9 @@ # / \ # a t # We say that "rgeat" is a scrambled string of "great". -# +# # Similarly, if we continue to swap the children of nodes "eat" and "at", it produces a scrambled string "rgtae". -# +# # rgtae # / \ # rg tae @@ -35,7 +35,7 @@ # / \ # t a # We say that "rgtae" is a scrambled string of "great". -# +# # Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1. # @@ -54,7 +54,7 @@ def isScramble(self, s1, s2): for j in xrange(len(s2)): if s1[i] == s2[j]: result[1][i][j] = True - + for n in xrange(2, len(s1) + 1): for i in xrange(len(s1) - n + 1): for j in xrange(len(s2) - n + 1): @@ -63,7 +63,7 @@ def isScramble(self, s1, s2): result[k][i][j + n - k] and result[n - k][i + k][j]: result[n][i][j] = True break - + return result[n][0][0] if __name__ == "__main__": diff --git a/Python/search-a-2d-matrix-ii.py b/Python/search-a-2d-matrix-ii.py index 3540dfd6c..e61290905 100644 --- a/Python/search-a-2d-matrix-ii.py +++ b/Python/search-a-2d-matrix-ii.py @@ -30,11 +30,11 @@ def searchMatrix(self, matrix, target): m = len(matrix) if m == 0: return False - + n = len(matrix[0]) if n == 0: return False - + i, j = 0, n - 1 while i < m and j >= 0: if matrix[i][j] == target: @@ -43,5 +43,5 @@ def searchMatrix(self, matrix, target): j -= 1 else: i += 1 - + return False diff --git a/Python/search-a-2d-matrix.py b/Python/search-a-2d-matrix.py index cc0e67e95..793894f2a 100644 --- a/Python/search-a-2d-matrix.py +++ b/Python/search-a-2d-matrix.py @@ -2,13 +2,13 @@ # Space: O(1) # # Write an efficient algorithm that searches for a value in an m x n matrix. This matrix has the following properties: -# +# # Integers in each row are sorted from left to right. # The first integer of each row is greater than the last integer of the previous row. # For example, -# +# # Consider the following matrix: -# +# # [ # [1, 3, 5, 7], # [10, 11, 16, 20], @@ -26,7 +26,7 @@ def searchMatrix(self, matrix, target): """ if not matrix: return False - + m, n = len(matrix), len(matrix[0]) left, right = 0, m * n while left < right: diff --git a/Python/search-for-a-range.py b/Python/search-for-a-range.py index 74bf58ac7..9d3ad9766 100644 --- a/Python/search-for-a-range.py +++ b/Python/search-for-a-range.py @@ -2,11 +2,11 @@ # Space: O(1) # # Given a sorted array of integers, find the starting and ending position of a given target value. -# +# # Your algorithm's runtime complexity must be in the order of O(log n). -# +# # If the target is not found in the array, return [-1, -1]. -# +# # For example, # Given [5, 7, 7, 8, 8, 10] and target value 8, # return [3, 4]. @@ -26,7 +26,7 @@ def searchRange(self, nums, target): # Find the first idx where nums[idx] > target right = self.binarySearch(lambda x, y: x > y, nums, target) return [left, right - 1] - + def binarySearch(self, compare, nums, target): left, right = 0, len(nums) while left < right: diff --git a/Python/search-in-rotated-sorted-array-ii.py b/Python/search-in-rotated-sorted-array-ii.py index 8649ec528..43a607f05 100644 --- a/Python/search-in-rotated-sorted-array-ii.py +++ b/Python/search-in-rotated-sorted-array-ii.py @@ -3,9 +3,9 @@ # # Follow up for "Search in Rotated Sorted Array": # What if duplicates are allowed? -# +# # Would this affect the run-time complexity? How and why? -# +# # Write a function to determine if a given target is in the array. # @@ -17,10 +17,10 @@ def search(self, nums, target): :rtype: int """ left, right = 0, len(nums) - 1 - + while left <= right: mid = left + (right - left) / 2 - + if nums[mid] == target: return True elif nums[mid] == nums[left]: @@ -32,7 +32,7 @@ def search(self, nums, target): left = mid + 1 return False - + if __name__ == "__main__": print Solution().search([3, 5, 1], 3) diff --git a/Python/search-in-rotated-sorted-array.py b/Python/search-in-rotated-sorted-array.py index d95d92382..4c93474b0 100644 --- a/Python/search-in-rotated-sorted-array.py +++ b/Python/search-in-rotated-sorted-array.py @@ -2,11 +2,11 @@ # Space: O(1) # # Suppose a sorted array is rotated at some pivot unknown to you beforehand. -# +# # (i.e., 0 1 2 4 5 6 7 might become 4 5 6 7 0 1 2). -# +# # You are given a target value to search. If found in the array return its index, otherwise return -1. -# +# # You may assume no duplicate exists in the array. # @@ -18,10 +18,10 @@ def search(self, nums, target): :rtype: int """ left, right = 0, len(nums) - 1 - + while left <= right: mid = left + (right - left) / 2 - + if nums[mid] == target: return mid elif (nums[mid] >= nums[left] and nums[left] <= target < nums[mid]) or \ @@ -31,7 +31,7 @@ def search(self, nums, target): left = mid + 1 return -1 - + if __name__ == "__main__": print Solution().search([3, 5, 1], 3) diff --git a/Python/search-insert-position.py b/Python/search-insert-position.py index a5be023ed..89b3f18de 100644 --- a/Python/search-insert-position.py +++ b/Python/search-insert-position.py @@ -2,11 +2,11 @@ # Space: O(1) # # Given a sorted array and a target value, return the index if the target is found. -# +# # If not, return the index where it would be if it were inserted in order. -# +# # You may assume no duplicates in the array. -# +# # Here are few examples. # [1,3,5,6], 5 -> 2 # [1,3,5,6], 2 -> 1 diff --git a/Python/self-crossing.py b/Python/self-crossing.py index cdb4439ab..76fd0d7d2 100644 --- a/Python/self-crossing.py +++ b/Python/self-crossing.py @@ -46,7 +46,7 @@ def isSelfCrossing(self, x): # 2 # 3 ┌────┐ # └─══>┘1 - # 4 0 (overlapped) + # 4 0 (overlapped) return True for i in xrange(3, len(x)): @@ -61,7 +61,7 @@ def isSelfCrossing(self, x): x[i - 1] <= x[i - 3] and x[i - 5] + x[i - 1] >= x[i - 3]: # Case 2: # i-4 - # ┌──┐ + # ┌──┐ # │i<┼─┐ # i-3│ i-5│i-1 # └────┘ diff --git a/Python/self-dividing-numbers.py b/Python/self-dividing-numbers.py index 84050e5a8..d3428690d 100644 --- a/Python/self-dividing-numbers.py +++ b/Python/self-dividing-numbers.py @@ -12,7 +12,7 @@ # including the bounds if possible. # # Example 1: -# Input: +# Input: # left = 1, right = 22 # Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22] # @@ -33,7 +33,7 @@ def isDividingNumber(num): return False n /= 10 return True - + result = [] for num in xrange(left, right+1): if isDividingNumber(num): diff --git a/Python/sentence-screen-fitting.py b/Python/sentence-screen-fitting.py index 91f45ca0c..eb5f98046 100644 --- a/Python/sentence-screen-fitting.py +++ b/Python/sentence-screen-fitting.py @@ -12,7 +12,7 @@ def wordsTyping(self, sentence, rows, cols): def words_fit(sentence, start, cols): if len(sentence[start]) > cols: return 0 - + s, count = len(sentence[start]), 1 i = (start + 1) % len(sentence) while s + 1 + len(sentence[i]) <= cols: diff --git a/Python/serialize-and-deserialize-binary-tree.py b/Python/serialize-and-deserialize-binary-tree.py index d69f43dd0..1806df30f 100644 --- a/Python/serialize-and-deserialize-binary-tree.py +++ b/Python/serialize-and-deserialize-binary-tree.py @@ -6,10 +6,10 @@ # or memory buffer, or transmitted across a network connection link # to be reconstructed later in the same or another computer environment. # -# Design an algorithm to serialize and deserialize a binary tree. +# Design an algorithm to serialize and deserialize a binary tree. # There is no restriction on how your serialization/deserialization -# algorithm should work. You just need to ensure that a binary tree can -# be serialized to a string and this string can be deserialized to the +# algorithm should work. You just need to ensure that a binary tree can +# be serialized to a string and this string can be deserialized to the # original tree structure. # # For example, you may serialize the following tree @@ -20,9 +20,9 @@ # / \ # 4 5 # as "[1,2,3,null,null,4,5]", just the same as how LeetCode OJ serializes -# a binary tree. You do not necessarily need to follow this format, so +# a binary tree. You do not necessarily need to follow this format, so # please be creative and come up with different approaches yourself. -# Note: Do not use class member/global/static variables to store states. +# Note: Do not use class member/global/static variables to store states. # Your serialize and deserialize algorithms should be stateless. # @@ -36,7 +36,7 @@ class Codec: def serialize(self, root): """Encodes a tree to a single string. - + :type root: TreeNode :rtype: str """ @@ -54,7 +54,7 @@ def serializeHelper(node): def deserialize(self, data): """Decodes your encoded data to tree. - + :type data: str :rtype: TreeNode """ diff --git a/Python/set-matrix-zeroes.py b/Python/set-matrix-zeroes.py index ea770632c..4a08ded96 100644 --- a/Python/set-matrix-zeroes.py +++ b/Python/set-matrix-zeroes.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place. -# +# # Follow up: # Did you use extra space? # A straight forward solution using O(mn) space is probably a bad idea. @@ -16,21 +16,21 @@ class Solution: def setZeroes(self, matrix): first_col = reduce(lambda acc, i: acc or matrix[i][0] == 0, xrange(len(matrix)), False) first_row = reduce(lambda acc, j: acc or matrix[0][j] == 0, xrange(len(matrix[0])), False) - + for i in xrange(1, len(matrix)): for j in xrange(1, len(matrix[0])): if matrix[i][j] == 0: matrix[i][0], matrix[0][j] = 0, 0 - + for i in xrange(1, len(matrix)): for j in xrange(1, len(matrix[0])): if matrix[i][0] == 0 or matrix[0][j] == 0: matrix[i][j] = 0 - + if first_col: for i in xrange(len(matrix)): matrix[i][0] = 0 - + if first_row: for j in xrange(len(matrix[0])): matrix[0][j] = 0 diff --git a/Python/set-mismatch.py b/Python/set-mismatch.py index 662ae25d4..d9a24acda 100644 --- a/Python/set-mismatch.py +++ b/Python/set-mismatch.py @@ -37,7 +37,7 @@ def findErrorNums(self, nums): # Time: O(n) -# Space: O(1) +# Space: O(1) class Solution2(object): def findErrorNums(self, nums): """ @@ -59,7 +59,7 @@ def findErrorNums(self, nums): # Time: O(n) -# Space: O(1) +# Space: O(1) class Solution3(object): def findErrorNums(self, nums): """ diff --git a/Python/shopping-offers.py b/Python/shopping-offers.py index 63fe142b0..743d8d0a3 100644 --- a/Python/shopping-offers.py +++ b/Python/shopping-offers.py @@ -19,24 +19,24 @@ # Example 1: # Input: [2,5], [[3,0,5],[1,2,10]], [3,2] # Output: 14 -# Explanation: -# There are two kinds of items, A and B. Their prices are $2 and $5 respectively. +# Explanation: +# There are two kinds of items, A and B. Their prices are $2 and $5 respectively. # In special offer 1, you can pay $5 for 3A and 0B -# In special offer 2, you can pay $10 for 1A and 2B. +# In special offer 2, you can pay $10 for 1A and 2B. # You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A. # Example 2: # Input: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1] # Output: 11 -# Explanation: -# The price of A is $2, and $3 for B, $4 for C. -# You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. -# You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. +# Explanation: +# The price of A is $2, and $3 for B, $4 for C. +# You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C. +# You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C. # You cannot add more items, though only $9 for 2A ,2B and 1C. # Note: # There are at most 6 kinds of items, 100 special offers. # For each item, you need to buy at most 6 of them. # You are not allowed to buy more items than you want, even if that would lower the overall price. - + class Solution(object): def shoppingOffers(self, price, special, needs): """ diff --git a/Python/shortest-distance-from-all-buildings.py b/Python/shortest-distance-from-all-buildings.py index 7b5fd2ae5..19a0b0082 100644 --- a/Python/shortest-distance-from-all-buildings.py +++ b/Python/shortest-distance-from-all-buildings.py @@ -10,7 +10,7 @@ def shortestDistance(self, grid): def bfs(grid, dists, cnts, x, y): dist, m, n = 0, len(grid), len(grid[0]) visited = [[False for _ in xrange(n)] for _ in xrange(m)] - + pre_level = [(x, y)] visited[x][y] = True while pre_level: @@ -24,7 +24,7 @@ def bfs(grid, dists, cnts, x, y): dists[I][J] += dist cur_level.append((I, J)) visited[I][J] = True - + pre_level = cur_level diff --git a/Python/shortest-palindrome.py b/Python/shortest-palindrome.py index f092ca4bf..e267e597e 100644 --- a/Python/shortest-palindrome.py +++ b/Python/shortest-palindrome.py @@ -1,8 +1,8 @@ # Time: O(n) # Space: O(n) # -# Given a string S, you are allowed to convert it to a palindrome -# by adding characters in front of it. Find and return the shortest +# Given a string S, you are allowed to convert it to a palindrome +# by adding characters in front of it. Find and return the shortest # palindrome you can find by performing this transformation. # # For example: @@ -32,7 +32,7 @@ def getPrefix(pattern): if not s: return s - + A = s + s[::-1] prefix = getPrefix(A) i = prefix[-1] @@ -60,7 +60,7 @@ def preProcess(s): return string string = preProcess(s) - palindrome = [0] * len(string) + palindrome = [0] * len(string) center, right = 0, 0 for i in xrange(1, len(string) - 1): i_mirror = 2 * center - i @@ -73,8 +73,8 @@ def preProcess(s): palindrome[i] += 1 if i + palindrome[i] > right: - center, right = i, i + palindrome[i] - + center, right = i, i + palindrome[i] + max_len = 0 for i in xrange(1, len(string) - 1): if i - palindrome[i] == 1: diff --git a/Python/shortest-unsorted-continuous-subarray.py b/Python/shortest-unsorted-continuous-subarray.py index 1de3ba3fe..637b692a0 100644 --- a/Python/shortest-unsorted-continuous-subarray.py +++ b/Python/shortest-unsorted-continuous-subarray.py @@ -31,7 +31,7 @@ def findUnsortedSubarray(self, nums): min_from_right = min(min_from_right, nums[n-1-i]) if nums[i] < max_from_left: right = i if nums[n-1-i] > min_from_right: left = n-1-i - + # Time: O(nlogn) # Space: O(n) diff --git a/Python/simplify-path.py b/Python/simplify-path.py index b55e73a11..532b544c1 100644 --- a/Python/simplify-path.py +++ b/Python/simplify-path.py @@ -2,12 +2,12 @@ # Space: O(n) # # Given an absolute path for a file (Unix-style), simplify it. -# +# # For example, # path = "/home/", => "/home" # path = "/a/./b/../../c/", => "/c" # click to show corner cases. -# +# # Corner Cases: # Did you consider the case where path = "/../"? # In this case, you should return "/". diff --git a/Python/single-element-in-a-sorted-array.py b/Python/single-element-in-a-sorted-array.py index 4a7dcb18b..ef54a8bef 100644 --- a/Python/single-element-in-a-sorted-array.py +++ b/Python/single-element-in-a-sorted-array.py @@ -12,7 +12,7 @@ # Input: [3,3,7,7,10,11,11] # Output: 10 # Note: Your solution should run in O(log n) time and O(1) space. - + class Solution(object): def singleNonDuplicate(self, nums): """ diff --git a/Python/single-number-ii.py b/Python/single-number-ii.py index e87276c96..586b80c30 100644 --- a/Python/single-number-ii.py +++ b/Python/single-number-ii.py @@ -1,8 +1,8 @@ # Time: O(n) # Space: O(1) -# +# # Given an array of integers, every element appears three times except for one. Find that single one. -# +# # Note: # Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? import collections @@ -16,7 +16,7 @@ def singleNumber(self, A): for x in A: one, two = (~x & one) | (x & ~one & ~two), (~x & two) | (x & one) return one - + class Solution2(object): # @param A, a list of integer # @return an integer diff --git a/Python/single-number.py b/Python/single-number.py index 88047ae25..742ea9c09 100644 --- a/Python/single-number.py +++ b/Python/single-number.py @@ -2,7 +2,7 @@ # Space: O(1) # # Given an array of integers, every element appears twice except for one. Find that single one. -# +# # Note: # Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory? # diff --git a/Python/sliding-window-maximum.py b/Python/sliding-window-maximum.py index eeaffeedb..1fb613e50 100644 --- a/Python/sliding-window-maximum.py +++ b/Python/sliding-window-maximum.py @@ -19,7 +19,7 @@ # 1 3 -1 -3 5 [3 6 7] 7 # Therefore, return the max sliding window as [3,3,5,5,6,7]. # -# Note: +# Note: # You may assume k is always valid, ie: 1 <= k <= input array's size for non-empty array. # # Follow up: diff --git a/Python/smallest-rotation-with-highest-score.py b/Python/smallest-rotation-with-highest-score.py index c46d7421d..c37570ea0 100644 --- a/Python/smallest-rotation-with-highest-score.py +++ b/Python/smallest-rotation-with-highest-score.py @@ -1,9 +1,9 @@ # Time: O(n) # Space: O(n) -# Given an array A, we may rotate it by a non-negative integer K +# Given an array A, we may rotate it by a non-negative integer K # so that the array becomes A[K], A[K+1], A{K+2], ... A[A.length - 1], A[0], A[1], ..., A[K-1]. -# Afterward, any entries that are less than or equal to their index are worth 1 point. +# Afterward, any entries that are less than or equal to their index are worth 1 point. # # For example, if we have [2, 4, 1, 3, 0], and we rotate by K = 2, # it becomes [1, 3, 0, 2, 4]. @@ -17,8 +17,8 @@ # Example 1: # Input: [2, 3, 1, 4, 0] # Output: 3 -# Explanation: -# Scores for each K are listed below: +# Explanation: +# Scores for each K are listed below: # K = 0, A = [2,3,1,4,0], score 2 # K = 1, A = [3,1,4,0,2], score 3 # K = 2, A = [1,4,0,2,3], score 3 @@ -35,7 +35,7 @@ # Note: # - A will have length at most 20000. # - A[i] will be in the range [0, A.length]. - + class Solution(object): def bestRotation(self, A): """ @@ -49,4 +49,4 @@ def bestRotation(self, A): for i in xrange(1, N): change[i] += change[i-1] return change.index(max(change)) - + diff --git a/Python/sort-colors.py b/Python/sort-colors.py index 5e42aa69b..9a7b2b8dd 100644 --- a/Python/sort-colors.py +++ b/Python/sort-colors.py @@ -1,21 +1,21 @@ # Time: O(n) # Space: O(1) # -# Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, +# Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, # with the colors in the order red, white and blue. -# +# # Here, we will use the integers 0, 1, and 2 to represent the color red, white, and blue respectively. -# +# # Note: # You are not suppose to use the library's sort function for this problem. -# +# # click to show follow up. -# +# # Follow up: # A rather straight forward solution is a two-pass algorithm using counting sort. -# First, iterate the array counting number of 0's, 1's, and 2's, +# First, iterate the array counting number of 0's, 1's, and 2's, # then overwrite array with total number of 0's, then 1's and followed by 2's. -# +# # Could you come up with an one-pass algorithm using only constant space? # @@ -27,7 +27,7 @@ def sortColors(self, nums): """ def triPartition(nums, target): i, j, n = 0, 0, len(nums) - 1 - + while j <= n: if nums[j] < target: nums[i], nums[j] = nums[j], nums[i] diff --git a/Python/sort-list.py b/Python/sort-list.py index 5427a88bd..d01ed7f56 100644 --- a/Python/sort-list.py +++ b/Python/sort-list.py @@ -9,43 +9,43 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, repr(self.next)) class Solution: # @param head, a ListNode - # @return a ListNode + # @return a ListNode def sortList(self, head): if head == None or head.next == None: return head - + fast, slow, prev = head, head, None while fast != None and fast.next != None: prev, fast, slow = slow, fast.next.next, slow.next prev.next = None - + sorted_l1 = self.sortList(head) sorted_l2 = self.sortList(slow) - + return self.mergeTwoLists(sorted_l1, sorted_l2) - + def mergeTwoLists(self, l1, l2): dummy = ListNode(0) - + cur = dummy while l1 != None and l2 != None: if l1.val <= l2.val: cur.next, cur, l1 = l1, l1, l1.next else: cur.next, cur, l2 = l2, l2, l2.next - + if l1 != None: cur.next = l1 if l2 != None: cur.next = l2 - + return dummy.next if __name__ == "__main__": diff --git a/Python/special-binary-string.py b/Python/special-binary-string.py index b743a60fa..ab8a7ced5 100644 --- a/Python/special-binary-string.py +++ b/Python/special-binary-string.py @@ -8,7 +8,7 @@ # Every prefix of the binary string has at least as many 1's as 0's. # Given a special string S, a move consists of choosing two consecutive, non-empty, # special substrings of S, and swapping them. -# (Two strings are consecutive if the last character of the first string is +# (Two strings are consecutive if the last character of the first string is # exactly one index before the first character of the second string.) # # At the end of any number of moves, what is the lexicographically largest resulting string possible? diff --git a/Python/spiral-matrix-ii.py b/Python/spiral-matrix-ii.py index d0b870bd5..d23da6b19 100644 --- a/Python/spiral-matrix-ii.py +++ b/Python/spiral-matrix-ii.py @@ -2,10 +2,10 @@ # Space: O(1) # # Given an integer n, generate a square matrix filled with elements from 1 to n2 in spiral order. -# +# # For example, # Given n = 3, -# +# # You should return the following matrix: # [ # [ 1, 2, 3 ], @@ -18,9 +18,9 @@ class Solution: # @return a list of lists of integer def generateMatrix(self, n): matrix = [[0 for _ in xrange(n)] for _ in xrange(n)] - + left, right, top, bottom, num = 0, n - 1, 0, n - 1, 1 - + while left <= right and top <= bottom: for j in xrange(left, right + 1): matrix[top][j] = num @@ -37,7 +37,7 @@ def generateMatrix(self, n): matrix[i][left] = num num += 1 left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1 - + return matrix diff --git a/Python/spiral-matrix.py b/Python/spiral-matrix.py index 5071ccc54..1e0734fe1 100644 --- a/Python/spiral-matrix.py +++ b/Python/spiral-matrix.py @@ -2,10 +2,10 @@ # Space: O(1) # # Given a matrix of m x n elements (m rows, n columns), return all elements of the matrix in spiral order. -# +# # For example, # Given the following matrix: -# +# # [ # [ 1, 2, 3 ], # [ 4, 5, 6 ], @@ -21,9 +21,9 @@ def spiralOrder(self, matrix): result = [] if matrix == []: return result - + left, right, top, bottom = 0, len(matrix[0]) - 1, 0, len(matrix) - 1 - + while left <= right and top <= bottom: for j in xrange(left, right + 1): result.append(matrix[top][j]) @@ -36,7 +36,7 @@ def spiralOrder(self, matrix): if left < right: result.append(matrix[i][left]) left, right, top, bottom = left + 1, right - 1, top + 1, bottom - 1 - + return result diff --git a/Python/split-array-into-consecutive-subsequences.py b/Python/split-array-into-consecutive-subsequences.py index c5c0008ca..2e6b113d5 100644 --- a/Python/split-array-into-consecutive-subsequences.py +++ b/Python/split-array-into-consecutive-subsequences.py @@ -9,14 +9,14 @@ # Input: [1,2,3,3,4,5] # Output: True # Explanation: -# You can split them into two consecutive subsequences : +# You can split them into two consecutive subsequences : # 1, 2, 3 # 3, 4, 5 # Example 2: # Input: [1,2,3,3,4,4,5,5] # Output: True # Explanation: -# You can split them into two consecutive subsequences : +# You can split them into two consecutive subsequences : # 1, 2, 3, 4, 5 # 3, 4, 5 # Example 3: diff --git a/Python/split-array-with-equal-sum.py b/Python/split-array-with-equal-sum.py index 036df2334..825e79418 100644 --- a/Python/split-array-with-equal-sum.py +++ b/Python/split-array-with-equal-sum.py @@ -9,7 +9,7 @@ def splitArray(self, nums): """ if len(nums) < 7: return False - + accumulated_sum = [0] * len(nums) accumulated_sum[0] = nums[0] for i in xrange(1, len(nums)): diff --git a/Python/split-array-with-same-average.py b/Python/split-array-with-same-average.py index aaca0d8bf..f4a83ec9f 100644 --- a/Python/split-array-with-same-average.py +++ b/Python/split-array-with-same-average.py @@ -8,7 +8,7 @@ # the average value of B is equal to the average value of C, and B and C are both non-empty. # # Example : -# Input: +# Input: # [1,2,3,4,5,6,7,8] # Output: true # Explanation: We can split the array into [1,4,5,8] and [2,3,6,7], and both of them have the average of 4.5. diff --git a/Python/split-concatenated-strings.py b/Python/split-concatenated-strings.py index 4693b532e..4ad1d593f 100644 --- a/Python/split-concatenated-strings.py +++ b/Python/split-concatenated-strings.py @@ -17,9 +17,9 @@ # Example: # Input: "abc", "xyz" # Output: "zyxcba" -# Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-", -# where '-' represents the looped status. -# The answer string came from the fourth looped one, +# Explanation: You can get the looped string "-abcxyz-", "-abczyx-", "-cbaxyz-", "-cbazyx-", +# where '-' represents the looped status. +# The answer string came from the fourth looped one, # where you could cut from the middle character 'a' and get "zyxcba". # Note: # The input strings will only contain lowercase letters. diff --git a/Python/split-linked-list-in-parts.py b/Python/split-linked-list-in-parts.py index 27b6e791f..4189510f1 100644 --- a/Python/split-linked-list-in-parts.py +++ b/Python/split-linked-list-in-parts.py @@ -15,7 +15,7 @@ # # Examples 1->2->3->4, k = 5 // 5 equal parts [ [1], [2], [3], [4], null ] # Example 1: -# Input: +# Input: # root = [1, 2, 3], k = 5 # Output: [[1],[2],[3],[],[]] # Explanation: @@ -26,7 +26,7 @@ # The last element output[4] is null, but it's string representation as a ListNode is []. # # Example 2: -# Input: +# Input: # root = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3 # Output: [[1, 2, 3, 4], [5, 6, 7], [8, 9, 10]] # Explanation: diff --git a/Python/sqrtx.py b/Python/sqrtx.py index 8ac7af2c8..05296b789 100644 --- a/Python/sqrtx.py +++ b/Python/sqrtx.py @@ -2,7 +2,7 @@ # Space: O(1) # Implement int sqrt(int x). -# +# # Compute and return the square root of x. class Solution(object): @@ -13,7 +13,7 @@ def mySqrt(self, x): """ if x < 2: return x - + left, right = 1, x // 2 while left <= right: mid = left + (right - left) // 2 @@ -27,4 +27,4 @@ def mySqrt(self, x): if __name__ == "__main__": print Solution().mySqrt(10) - + diff --git a/Python/squirrel-simulation.py b/Python/squirrel-simulation.py index 11935ae5e..f0057eec2 100644 --- a/Python/squirrel-simulation.py +++ b/Python/squirrel-simulation.py @@ -13,7 +13,7 @@ def minDistance(self, height, width, tree, squirrel, nuts): """ def distance(a, b): return abs(a[0] - b[0]) + abs(a[1] - b[1]) - + result = 0 d = float("inf") for nut in nuts: diff --git a/Python/strange-printer.py b/Python/strange-printer.py index 534463942..811933aaf 100644 --- a/Python/strange-printer.py +++ b/Python/strange-printer.py @@ -17,11 +17,11 @@ # Example 2: # Input: "aba" # Output: 2 -# Explanation: Print "aaa" first and then print "b" from +# Explanation: Print "aaa" first and then print "b" from # the second place of the string, which will cover the existing character 'a'. # # Hint: Length of the given string will not exceed 100. - + class Solution(object): def strangePrinter(self, s): """ diff --git a/Python/string-to-integer-atoi.py b/Python/string-to-integer-atoi.py index 56b6c78e8..d59faea1a 100644 --- a/Python/string-to-integer-atoi.py +++ b/Python/string-to-integer-atoi.py @@ -2,27 +2,27 @@ # Space: O(1) # # Implement atoi to convert a string to an integer. -# -# Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below +# +# Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below # and ask yourself what are the possible input cases. -# +# # Notes: It is intended for this problem to be specified vaguely (ie, no given input specs). # You are responsible to gather all the input requirements up front. -# +# # spoilers alert... click to show requirements for atoi. -# +# # Requirements for atoi: -# The function first discards as many whitespace characters as necessary -# until the first non-whitespace character is found. Then, starting from this character, +# The function first discards as many whitespace characters as necessary +# until the first non-whitespace character is found. Then, starting from this character, # takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value. -# -# The string can contain additional characters after those that +# +# The string can contain additional characters after those that # form the integral number, which are ignored and have no effect on the behavior of this function. -# -# If the first sequence of non-whitespace characters in str is not a valid integral number, +# +# If the first sequence of non-whitespace characters in str is not a valid integral number, # or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed. -# -# If no valid conversion could be performed, a zero value is returned. +# +# If no valid conversion could be performed, a zero value is returned. # If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned. # @@ -35,14 +35,14 @@ def myAtoi(self, str): INT_MAX = 2147483647 INT_MIN = -2147483648 result = 0 - + if not str: return result - + i = 0 while i < len(str) and str[i].isspace(): i += 1 - + sign = 1 if str[i] == "+": i += 1 @@ -55,13 +55,13 @@ def myAtoi(self, str): return INT_MAX if sign > 0 else INT_MIN result = result * 10 + int(str[i]) i += 1 - + return sign * result if __name__ == "__main__": - print Solution().atoi("") + print Solution().atoi("") print Solution().atoi("-1") - print Solution().atoi("2147483647") - print Solution().atoi("2147483648") - print Solution().atoi("-2147483648") - print Solution().atoi("-2147483649") + print Solution().atoi("2147483647") + print Solution().atoi("2147483648") + print Solution().atoi("-2147483648") + print Solution().atoi("-2147483649") diff --git a/Python/strobogrammatic-number-ii.py b/Python/strobogrammatic-number-ii.py index da089c481..a914402b2 100644 --- a/Python/strobogrammatic-number-ii.py +++ b/Python/strobogrammatic-number-ii.py @@ -14,7 +14,7 @@ def findStrobogrammaticRecu(self, n, k): return [''] elif k == 1: return ['0', '1', '8'] - + result = [] for num in self.findStrobogrammaticRecu(n, k - 2): for key, val in self.lookup.iteritems(): diff --git a/Python/strobogrammatic-number-iii.py b/Python/strobogrammatic-number-iii.py index d45aa3d49..23bf9b7b1 100644 --- a/Python/strobogrammatic-number-iii.py +++ b/Python/strobogrammatic-number-iii.py @@ -25,7 +25,7 @@ def countStrobogrammaticUntil(self, num, can_start_with_0): count += 1 self.cache[num] = count return count - + for key, val in self.lookup.iteritems(): if can_start_with_0 or key != '0': if num[0] > key: @@ -36,7 +36,7 @@ def countStrobogrammaticUntil(self, num, can_start_with_0): elif num[0] == key: if len(num) == 2: # num is like 12". if num[-1] >= val: - count += 1 + count += 1 else: if num[-1] >= val: # num is like "102". count += self.countStrobogrammaticUntil(self.getMid(num), True); diff --git a/Python/strong-password-checker.py b/Python/strong-password-checker.py index 90269f9fc..2776c8dfa 100644 --- a/Python/strong-password-checker.py +++ b/Python/strong-password-checker.py @@ -36,7 +36,7 @@ def strongPasswordChecker(self, s): while i < len(s) and s[i] == s[i-1]: length += 1 i += 1 - + total_change_cnt += length / 3 if length % 3 == 0: one_change_cnt += 1 @@ -46,16 +46,16 @@ def strongPasswordChecker(self, s): three_change_cnt += 1 else: i += 1 - + if len(s) < 6: return max(missing_type_cnt, 6 - len(s)) elif len(s) <= 20: return max(missing_type_cnt, total_change_cnt) else: delete_cnt = len(s) - 20 - + total_change_cnt -= min(delete_cnt, one_change_cnt * 1) / 1 total_change_cnt -= min(max(delete_cnt - one_change_cnt, 0), two_change_cnt * 2) / 2 total_change_cnt -= min(max(delete_cnt - one_change_cnt - 2 * two_change_cnt, 0), three_change_cnt * 3) / 3 - + return delete_cnt + max(missing_type_cnt, total_change_cnt) diff --git a/Python/student-attendance-record-i.py b/Python/student-attendance-record-i.py index 07c836558..e63511eae 100644 --- a/Python/student-attendance-record-i.py +++ b/Python/student-attendance-record-i.py @@ -34,4 +34,4 @@ def checkRecord(self, s): if i < len(s) - 2 and s[i] == s[i+1] == s[i+2] == 'L': return False return True - + diff --git a/Python/student-attendance-record-ii.py b/Python/student-attendance-record-ii.py index 2b1cc99b6..c52793b45 100644 --- a/Python/student-attendance-record-ii.py +++ b/Python/student-attendance-record-ii.py @@ -14,11 +14,11 @@ # # Example 1: # Input: n = 2 -# Output: 8 +# Output: 8 # Explanation: # There are 8 records with length 2 will be regarded as rewardable: # "PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL" -# Only "AA" won't be regarded as rewardable owing to more than one absent times. +# Only "AA" won't be regarded as rewardable owing to more than one absent times. # Note: The value of n won't exceed 100,000. class Solution(object): diff --git a/Python/subarray-product-less-than-k.py b/Python/subarray-product-less-than-k.py index 3650ab757..9d947ca72 100644 --- a/Python/subarray-product-less-than-k.py +++ b/Python/subarray-product-less-than-k.py @@ -35,4 +35,4 @@ def numSubarrayProductLessThanK(self, nums, k): start += 1 result += i-start+1 return result - + diff --git a/Python/subsets-ii.py b/Python/subsets-ii.py index f30e20486..ebd20f0bb 100644 --- a/Python/subsets-ii.py +++ b/Python/subsets-ii.py @@ -2,13 +2,13 @@ # Space: O(1) # Given a collection of integers that might contain duplicates, S, return all possible subsets. -# +# # Note: # Elements in a subset must be in non-descending order. # The solution set must not contain duplicate subsets. # For example, # If S = [1,2,2], a solution is: -# +# # [ # [2], # [1], @@ -49,7 +49,7 @@ def subsetsWithDup(self, nums): result = [] i, count = 0, 1 << len(nums) nums.sort() - + while i < count: cur = [] for j in xrange(len(nums)): @@ -58,7 +58,7 @@ def subsetsWithDup(self, nums): if cur not in result: result.append(cur) i += 1 - + return result @@ -73,7 +73,7 @@ def subsetsWithDup(self, nums): result = [] self.subsetsWithDupRecu(result, [], sorted(nums)) return result - + def subsetsWithDupRecu(self, result, cur, nums): if not nums: if cur not in result: @@ -82,6 +82,6 @@ def subsetsWithDupRecu(self, result, cur, nums): self.subsetsWithDupRecu(result, cur, nums[1:]) self.subsetsWithDupRecu(result, cur + [nums[0]], nums[1:]) - + if __name__ == "__main__": print Solution().subsetsWithDup([1, 2, 2]) diff --git a/Python/subsets.py b/Python/subsets.py index 113b810e7..83ba120db 100644 --- a/Python/subsets.py +++ b/Python/subsets.py @@ -2,13 +2,13 @@ # Space: O(1) # Given a set of distinct integers, S, return all possible subsets. -# +# # Note: # Elements in a subset must be in non-descending order. # The solution set must not contain duplicate subsets. # For example, # If S = [1,2,3], a solution is: -# +# # [ # [3], # [1], @@ -47,7 +47,7 @@ def subsets(self, nums): result = [] i, count = 0, 1 << len(nums) nums.sort() - + while i < count: cur = [] for j in xrange(len(nums)): @@ -55,7 +55,7 @@ def subsets(self, nums): cur.append(nums[j]) result.append(cur) i += 1 - + return result @@ -68,11 +68,11 @@ def subsets(self, nums): :rtype: List[List[int]] """ return self.subsetsRecu([], sorted(nums)) - + def subsetsRecu(self, cur, nums): if not nums: return [cur] - + return self.subsetsRecu(cur, nums[1:]) + self.subsetsRecu(cur + [nums[0]], nums[1:]) diff --git a/Python/subtree-of-another-tree.py b/Python/subtree-of-another-tree.py index 9b0987f33..423941776 100644 --- a/Python/subtree-of-another-tree.py +++ b/Python/subtree-of-another-tree.py @@ -16,7 +16,7 @@ # / \ # 1 2 # Given tree t: -# 4 +# 4 # / \ # 1 2 # Return true, because t has the same structure and node values with a subtree of s. @@ -64,5 +64,5 @@ def preOrderTraverse(s, t): (isSame(s, t) or \ preOrderTraverse(s.left, t) or \ preOrderTraverse(s.right, t)) - + return preOrderTraverse(s, t) diff --git a/Python/sudoku-solver.py b/Python/sudoku-solver.py index fff654616..39835b421 100644 --- a/Python/sudoku-solver.py +++ b/Python/sudoku-solver.py @@ -2,9 +2,9 @@ # Space: (1) # # Write a program to solve a Sudoku puzzle by filling the empty cells. -# +# # Empty cells are indicated by the character '.'. -# +# # You may assume that there will be only one unique solution. # @@ -29,7 +29,7 @@ def isValid(board, x, y): j += 1 i += 1 return True - + def solver(board): for i in xrange(len(board)): for j in xrange(len(board[0])): diff --git a/Python/sum-of-left-leaves.py b/Python/sum-of-left-leaves.py index 4ca0bd5c0..fb0787415 100644 --- a/Python/sum-of-left-leaves.py +++ b/Python/sum-of-left-leaves.py @@ -34,5 +34,5 @@ def sumOfLeftLeavesHelper(root, is_left): return root.val if is_left else 0 return sumOfLeftLeavesHelper(root.left, True) + \ sumOfLeftLeavesHelper(root.right, False) - + return sumOfLeftLeavesHelper(root, False) diff --git a/Python/sum-root-to-leaf-numbers.py b/Python/sum-root-to-leaf-numbers.py index 775c1920b..faad48bff 100644 --- a/Python/sum-root-to-leaf-numbers.py +++ b/Python/sum-root-to-leaf-numbers.py @@ -2,19 +2,19 @@ # Space: O(h), h is height of binary tree # # Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number. -# +# # An example is the root-to-leaf path 1->2->3 which represents the number 123. -# +# # Find the total sum of all root-to-leaf numbers. -# +# # For example, -# +# # 1 # / \ # 2 3 # The root-to-leaf path 1->2 represents the number 12. # The root-to-leaf path 1->3 represents the number 13. -# +# # Return the sum = 12 + 13 = 25. # @@ -30,18 +30,18 @@ class Solution: # @return an integer def sumNumbers(self, root): return self.sumNumbersRecu(root, 0) - + def sumNumbersRecu(self, root, num): if root is None: return 0 - + if root.left is None and root.right is None: return num * 10 + root.val - + return self.sumNumbersRecu(root.left, num * 10 + root.val) + self.sumNumbersRecu(root.right, num * 10 + root.val) if __name__ == "__main__": root = TreeNode(1) root.left = TreeNode(2) root.right = TreeNode(3) - print Solution().sumNumbers(root) + print Solution().sumNumbers(root) diff --git a/Python/super-washing-machines.py b/Python/super-washing-machines.py index 1db926366..202cba8db 100644 --- a/Python/super-washing-machines.py +++ b/Python/super-washing-machines.py @@ -20,27 +20,27 @@ # # Output: 3 # -# Explanation: +# Explanation: # 1st move: 1 0 <-- 5 => 1 1 4 -# 2nd move: 1 <-- 1 <-- 4 => 2 1 3 -# 3rd move: 2 1 <-- 3 => 2 2 2 +# 2nd move: 1 <-- 1 <-- 4 => 2 1 3 +# 3rd move: 2 1 <-- 3 => 2 2 2 # Example2 # # Input: [0,3,0] # # Output: 2 # -# Explanation: -# 1st move: 0 <-- 3 0 => 1 2 0 -# 2nd move: 1 2 --> 0 => 1 1 1 +# Explanation: +# 1st move: 0 <-- 3 0 => 1 2 0 +# 2nd move: 1 2 --> 0 => 1 1 1 # Example3 # # Input: [0,2,0] # # Output: -1 # -# Explanation: -# It's impossible to make all the three washing machines have the same number of dresses. +# Explanation: +# It's impossible to make all the three washing machines have the same number of dresses. # Note: # The range of n is [1, 10000]. # The range of dresses number in a super washing machine is [0, 1e5]. @@ -54,7 +54,7 @@ def findMinMoves(self, machines): total = sum(machines) if total % len(machines): return -1 - result, target, curr = 0, total / len(machines), 0 + result, target, curr = 0, total / len(machines), 0 for n in machines: curr += n - target result = max(result, max(n - target, abs(curr))) diff --git a/Python/swap-nodes-in-pairs.py b/Python/swap-nodes-in-pairs.py index c11df1225..91c039f8f 100644 --- a/Python/swap-nodes-in-pairs.py +++ b/Python/swap-nodes-in-pairs.py @@ -2,10 +2,10 @@ # Space: O(1) # # Given a linked list, swap every two adjacent nodes and return its head. -# +# # For example, # Given 1->2->3->4, you should return the list as 2->1->4->3. -# +# # Your algorithm should use only constant space. # You may not modify the values in the list, only nodes itself can be changed. # @@ -15,7 +15,7 @@ class ListNode: def __init__(self, x): self.val = x self.next = None - + def __repr__(self): if self: return "{} -> {}".format(self.val, self.next) @@ -39,4 +39,3 @@ def swapPairs(self, head): head = ListNode(1) head.next, head.next.next, head.next.next.next = ListNode(2), ListNode(3), ListNode(4) print Solution().swapPairs(head) - \ No newline at end of file diff --git a/Python/swim-in-rising-water.py b/Python/swim-in-rising-water.py index 0418afd34..f2034acac 100644 --- a/Python/swim-in-rising-water.py +++ b/Python/swim-in-rising-water.py @@ -54,7 +54,7 @@ def union_set(self, x, y): return False self.set[min(x_root, y_root)] = max(x_root, y_root) return True - + class Solution(object): def swimInWater(self, grid): @@ -68,7 +68,7 @@ def swimInWater(self, grid): for j in xrange(n): positions[grid[i][j]] = (i, j) directions = ((-1, 0), (1, 0), (0, -1), (0, 1)) - + union_find = UnionFind(n**2) for elevation in xrange(n**2): i, j = positions[elevation] @@ -79,4 +79,4 @@ def swimInWater(self, grid): if union_find.find_set(0) == union_find.find_set(n**2-1): return elevation return n**2-1 - + diff --git a/Python/symmetric-tree.py b/Python/symmetric-tree.py index 709864217..3bc99e50b 100644 --- a/Python/symmetric-tree.py +++ b/Python/symmetric-tree.py @@ -1,9 +1,9 @@ # Time: O(n) # Space: O(h), h is height of binary tree # Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). -# +# # For example, this binary tree is symmetric: -# +# # 1 # / \ # 2 2 @@ -36,24 +36,24 @@ def isSymmetric(self, root): stack = [] stack.append(root.left) stack.append(root.right) - + while stack: p, q = stack.pop(), stack.pop() - + if p is None and q is None: continue - + if p is None or q is None or p.val != q.val: return False - + stack.append(p.left) stack.append(q.right) - + stack.append(p.right) stack.append(q.left) - + return True - + # Recursive solution class Solution2: # @param root, a tree node @@ -61,9 +61,9 @@ class Solution2: def isSymmetric(self, root): if root is None: return True - + return self.isSymmetricRecu(root.left, root.right) - + def isSymmetricRecu(self, left, right): if left is None and right is None: return True @@ -77,4 +77,4 @@ def isSymmetricRecu(self, left, right): root.left.left, root.right.right = TreeNode(3), TreeNode(3) root.left.right, root.right.left = TreeNode(4), TreeNode(4) print Solution().isSymmetric(root) - + diff --git a/Python/tag-validator.py b/Python/tag-validator.py index 25946efa0..7409ec413 100644 --- a/Python/tag-validator.py +++ b/Python/tag-validator.py @@ -10,7 +10,7 @@ # TAG_CONTENT. Among them, is the start tag, # and is the end tag. The TAG_NAME in start and end tags should be the same. # A closed tag is valid if and only if the TAG_NAME and TAG_CONTENT are valid. -# 3. A valid TAG_NAME only contain upper-case letters, and has length in range [1,9]. +# 3. A valid TAG_NAME only contain upper-case letters, and has length in range [1,9]. # Otherwise, the TAG_NAME is invalid. # 4. A valid TAG_CONTENT may contain other valid closed tags, # cdata and any characters (see note1) EXCEPT unmatched <, @@ -31,9 +31,9 @@ # Valid Code Examples: # Input: "
This is the first line ]]>
" # Output: True -# Explanation: -# The code is wrapped in a closed tag :
and
. -# The TAG_NAME is valid, the TAG_CONTENT consists of some characters and cdata. +# Explanation: +# The code is wrapped in a closed tag :
and
. +# The TAG_NAME is valid, the TAG_CONTENT consists of some characters and cdata. # Although CDATA_CONTENT has unmatched start tag with invalid TAG_NAME, # it should be considered as plain text, not parsed as tag. # So TAG_CONTENT is valid, and then the code is valid. Thus return true. @@ -124,10 +124,10 @@ def validTag(s, i): if not tag: return False, i j = parseContent(s, j) - k = j + len(tag) + 2 + k = j + len(tag) + 2 if k >= len(s) or s[j:k+1] != "": return False, i return True, k+1 - + result, i = validTag(code, 0) return result and i == len(code) diff --git a/Python/teemo-attacking.py b/Python/teemo-attacking.py index 4ae1d8435..918242f76 100644 --- a/Python/teemo-attacking.py +++ b/Python/teemo-attacking.py @@ -11,18 +11,18 @@ # Example 1: # Input: [1,4], 2 # Output: 4 -# Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned immediately. -# This poisoned status will last 2 seconds until the end of time point 2. -# And at time point 4, Teemo attacks Ashe again, and causes Ashe to be in poisoned status for another 2 seconds. +# Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned immediately. +# This poisoned status will last 2 seconds until the end of time point 2. +# And at time point 4, Teemo attacks Ashe again, and causes Ashe to be in poisoned status for another 2 seconds. # So you finally need to output 4. # Example 2: # Input: [1,2], 2 # Output: 3 -# Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned. -# This poisoned status will last 2 seconds until the end of time point 2. -# However, at the beginning of time point 2, Teemo attacks Ashe again who is already in poisoned status. +# Explanation: At time point 1, Teemo starts attacking Ashe and makes Ashe be poisoned. +# This poisoned status will last 2 seconds until the end of time point 2. +# However, at the beginning of time point 2, Teemo attacks Ashe again who is already in poisoned status. # Since the poisoned status won't add up together, though the second poisoning attack will still work at time point 2, -# it will stop at the end of time point 3. +# it will stop at the end of time point 3. # So you finally need to output 3. # Note: # You may assume the length of given time series array won't exceed 10000. diff --git a/Python/ternary-expression-parser.py b/Python/ternary-expression-parser.py index f8928be3e..3f6e9b14b 100644 --- a/Python/ternary-expression-parser.py +++ b/Python/ternary-expression-parser.py @@ -17,7 +17,7 @@ def parseTernary(self, expression): first = stack.pop() stack.pop() # pop ':' second = stack.pop() - + if c == 'T': stack.append(first) else: @@ -25,5 +25,5 @@ def parseTernary(self, expression): else: stack.append(c) - + return str(stack[-1]) diff --git a/Python/text-justification.py b/Python/text-justification.py index b292c30f7..ccc5ab439 100644 --- a/Python/text-justification.py +++ b/Python/text-justification.py @@ -3,22 +3,22 @@ # # Given an array of words and a length L, format the text such that # each line has exactly L characters and is fully (left and right) justified. -# -# You should pack your words in a greedy approach; that is, pack -# as many words as you can in each line. Pad extra spaces ' ' +# +# You should pack your words in a greedy approach; that is, pack +# as many words as you can in each line. Pad extra spaces ' ' # when necessary so that each line has exactly L characters. -# -# Extra spaces between words should be distributed as evenly as possible. -# If the number of spaces on a line do not divide evenly between words, +# +# Extra spaces between words should be distributed as evenly as possible. +# If the number of spaces on a line do not divide evenly between words, # the empty slots on the left will be assigned more spaces than the slots on the right. -# +# # For the last line of text, it should be left justified and no extra space # is inserted between words. -# +# # For example, # words: ["This", "is", "an", "example", "of", "text", "justification."] # L: 16. -# +# # Return the formatted lines as: # [ # "This is an", diff --git a/Python/the-skyline-problem.py b/Python/the-skyline-problem.py index d7e17a89c..e032ac3e6 100644 --- a/Python/the-skyline-problem.py +++ b/Python/the-skyline-problem.py @@ -1,33 +1,33 @@ # Time: O(nlogn) # Space: O(n) # -# A city's skyline is the outer contour of the silhouette formed -# by all the buildings in that city when viewed from a distance. -# Now suppose you are given the locations and height of all the +# A city's skyline is the outer contour of the silhouette formed +# by all the buildings in that city when viewed from a distance. +# Now suppose you are given the locations and height of all the # buildings as shown on a cityscape photo (Figure A), write a -# program to output the skyline formed by these buildings +# program to output the skyline formed by these buildings # collectively (Figure B). -# -# The geometric information of each building is represented by a -# triplet of integers [Li, Ri, Hi], where Li and Ri are the x -# coordinates of the left and right edge of the ith building, -# respectively, and Hi is its height. It is guaranteed that 0 <= Li, -# Ri <= INT_MAX, 0 < Hi <= INT_MAX, and Ri - Li > 0. You may assume -# all buildings are perfect rectangles grounded on an absolutely +# +# The geometric information of each building is represented by a +# triplet of integers [Li, Ri, Hi], where Li and Ri are the x +# coordinates of the left and right edge of the ith building, +# respectively, and Hi is its height. It is guaranteed that 0 <= Li, +# Ri <= INT_MAX, 0 < Hi <= INT_MAX, and Ri - Li > 0. You may assume +# all buildings are perfect rectangles grounded on an absolutely # flat surface at height 0. -# +# # Notes: -# -# The number of buildings in any input list is guaranteed to be +# +# The number of buildings in any input list is guaranteed to be # in the range [0, 10000]. -# The input list is already sorted in ascending order by the +# The input list is already sorted in ascending order by the # left x position Li. # The output list must be sorted by the x position. -# There must be no consecutive horizontal lines of equal height -# in the output skyline. -# For instance, [...[2 3], [4 5], [7 5], [11 5], [12 7]...] is -# not acceptable; -# the three lines of height 5 should be merged into one +# There must be no consecutive horizontal lines of equal height +# in the output skyline. +# For instance, [...[2 3], [4 5], [7 5], [11 5], [12 7]...] is +# not acceptable; +# the three lines of height 5 should be merged into one # in the final output as such: [...[2 3], [4 5], [12 7], ...] # @@ -38,7 +38,7 @@ class Solution: # @return {integer[][]} def getSkyline(self, buildings): intervals = self.ComputeSkylineInInterval(buildings, 0, len(buildings)) - + res = [] last_end = -1 for interval in intervals: @@ -48,9 +48,9 @@ def getSkyline(self, buildings): last_end = interval[end] if last_end != -1: res.append([last_end, 0]) - + return res - + # Divide and Conquer. def ComputeSkylineInInterval(self, buildings, left_endpoint, right_endpoint): if right_endpoint - left_endpoint <= 1: @@ -59,12 +59,12 @@ def ComputeSkylineInInterval(self, buildings, left_endpoint, right_endpoint): left_skyline = self.ComputeSkylineInInterval(buildings, left_endpoint, mid) right_skyline = self.ComputeSkylineInInterval(buildings, mid, right_endpoint) return self.MergeSkylines(left_skyline, right_skyline) - + # Merge Sort. def MergeSkylines(self, left_skyline, right_skyline): i, j = 0, 0 merged = [] - + while i < len(left_skyline) and j < len(right_skyline): if left_skyline[i][end] < right_skyline[j][start]: merged.append(left_skyline[i]) @@ -78,12 +78,12 @@ def MergeSkylines(self, left_skyline, right_skyline): else: # left_skyline[i][start] > right_skyline[j][start]. j, i = self.MergeIntersectSkylines(merged, right_skyline[j], j, \ left_skyline[i], i) - + # Insert the remaining skylines. merged += left_skyline[i:] merged += right_skyline[j:] return merged - + # a[start] <= b[start] def MergeIntersectSkylines(self, merged, a, a_idx, b, b_idx): if a[end] <= b[end]: diff --git a/Python/total-hamming-distance.py b/Python/total-hamming-distance.py index fac26b25b..411be8f44 100644 --- a/Python/total-hamming-distance.py +++ b/Python/total-hamming-distance.py @@ -31,4 +31,4 @@ def totalHammingDistance(self, nums): counts[(num >> i) & 1] += 1 result += counts[0] * counts[1] return result - + diff --git a/Python/trapping-rain-water-ii.py b/Python/trapping-rain-water-ii.py index 8a1037979..8408de5f1 100644 --- a/Python/trapping-rain-water-ii.py +++ b/Python/trapping-rain-water-ii.py @@ -34,7 +34,7 @@ def trapRainWater(self, heightMap): return 0 is_visited = [[False for i in xrange(n)] for j in xrange(m)] - + heap = [] for i in xrange(m): heappush(heap, [heightMap[i][0], i, 0]) diff --git a/Python/trapping-rain-water.py b/Python/trapping-rain-water.py index c2d030382..28a0a7e41 100644 --- a/Python/trapping-rain-water.py +++ b/Python/trapping-rain-water.py @@ -3,11 +3,11 @@ # # Given n non-negative integers representing an elevation map where the width of each bar is 1, # compute how much water it is able to trap after raining. -# -# For example, +# +# For example, # Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. -# -# The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. +# +# The above elevation map is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. # In this case, 6 units of rain water (blue section) are being trapped. # @@ -20,21 +20,21 @@ def trap(self, A): for i in xrange(len(A)): if A[top] < A[i]: top = i - + second_top = 0 for i in xrange(top): if A[second_top] < A[i]: second_top = i result += A[second_top] - A[i] - + second_top = len(A) - 1 for i in reversed(xrange(top, len(A))): if A[second_top] < A[i]: second_top = i result += A[second_top] - A[i] - + return result - + # Time: O(n) # Space: O(n) class Solution2: @@ -43,20 +43,20 @@ class Solution2: def trap(self, A): result = 0 stack = [] - + for i in xrange(len(A)): mid_height = 0 while stack: [pos, height] = stack.pop() result += (min(height, A[i]) - mid_height) * (i - pos - 1) mid_height = height - + if A[i] < height: stack.append([pos, height]) break stack.append([i, A[i]]) - + return result - + if __name__ == "__main__": print Solution().trap([0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1]) diff --git a/Python/triangle.py b/Python/triangle.py index e05906104..7d2b541f6 100644 --- a/Python/triangle.py +++ b/Python/triangle.py @@ -2,7 +2,7 @@ # Space: O(n) # # Given a triangle, find the minimum path sum from top to bottom. Each step you may move to adjacent numbers on the row below. -# +# # For example, given the following triangle # [ # [2], @@ -11,7 +11,7 @@ # [4,1,8,3] # ] # The minimum path sum from top to bottom is 11 (i.e., 2 + 3 + 5 + 1 = 11). -# +# # Note: # Bonus point if you are able to do this using only O(n) extra space, where n is the total number of rows in the triangle. # @@ -22,7 +22,7 @@ class Solution: def minimumTotal(self, triangle): if not triangle: return 0 - + cur = triangle[0] + [float("inf")] for i in xrange(1, len(triangle)): next = [] @@ -30,9 +30,9 @@ def minimumTotal(self, triangle): for j in xrange(1, i + 1): next.append(triangle[i][j] + min(cur[j - 1], cur[j])) cur = next + [float("inf")] - + return reduce(min, cur) if __name__ == "__main__": print Solution().minimumTotal([[-1], [2, 3], [1, -1, -3]]) - + diff --git a/Python/trim-a-binary-search-tree.py b/Python/trim-a-binary-search-tree.py index 992b05378..e0f1c4c07 100644 --- a/Python/trim-a-binary-search-tree.py +++ b/Python/trim-a-binary-search-tree.py @@ -7,7 +7,7 @@ # return the new root of the trimmed binary search tree. # # Example 1: -# Input: +# Input: # 1 # / \ # 0 2 @@ -15,12 +15,12 @@ # L = 1 # R = 2 # -# Output: +# Output: # 1 # \ # 2 # Example 2: -# Input: +# Input: # 3 # / \ # 0 4 @@ -32,10 +32,10 @@ # L = 1 # R = 3 # -# Output: +# Output: # 3 -# / -# 2 +# / +# 2 # / # 1 @@ -62,4 +62,4 @@ def trimBST(self, root, L, R): return self.trimBST(root.left, L, R) root.left, root.right = self.trimBST(root.left, L, R), self.trimBST(root.right, L, R) return root - + diff --git a/Python/two-sum-ii-input-array-is-sorted.py b/Python/two-sum-ii-input-array-is-sorted.py index 247af6397..8ac26f965 100644 --- a/Python/two-sum-ii-input-array-is-sorted.py +++ b/Python/two-sum-ii-input-array-is-sorted.py @@ -1,15 +1,15 @@ # Time: O(n) # Space: O(1) # -# Given an array of integers that is already sorted in ascending order, +# Given an array of integers that is already sorted in ascending order, # find two numbers such that they add up to a specific target number. -# -# The function twoSum should return indices of the two numbers such that -# they add up to the target, where index1 must be less than index2. +# +# The function twoSum should return indices of the two numbers such that +# they add up to the target, where index1 must be less than index2. # Please note that your returned answers (both index1 and index2) are not zero-based. -# +# # You may assume that each input would have exactly one solution. -# +# # Input: numbers={2, 7, 11, 15}, target=9 # Output: index1=1, index2=2 # @@ -17,7 +17,7 @@ class Solution: def twoSum(self, nums, target): start, end = 0, len(nums) - 1 - + while start != end: sum = nums[start] + nums[end] if sum > target: diff --git a/Python/two-sum-iii-data-structure-design.py b/Python/two-sum-iii-data-structure-design.py index 808c80d91..05bf1c555 100644 --- a/Python/two-sum-iii-data-structure-design.py +++ b/Python/two-sum-iii-data-structure-design.py @@ -2,10 +2,10 @@ # Space: O(n) # Design and implement a TwoSum class. It should support the following operations: add and find. -# +# # add - Add the number to an internal data structure. # find - Find if there exists any pair of numbers which sum is equal to the value. -# +# # For example, # add(1); add(3); add(5); # find(4) -> true @@ -20,7 +20,7 @@ def __init__(self): initialize your data structure here """ self.lookup = defaultdict(int) - + def add(self, number): @@ -46,10 +46,10 @@ def find(self, value): if __name__ == "__main__": Sol = TwoSum() - + for i in (1, 3, 5): Sol.add(i) - + for i in (4, 7): print Sol.find(i) - + diff --git a/Python/two-sum-iv-input-is-a-bst.py b/Python/two-sum-iv-input-is-a-bst.py index e55b985bc..47a3b44d7 100644 --- a/Python/two-sum-iv-input-is-a-bst.py +++ b/Python/two-sum-iv-input-is-a-bst.py @@ -5,7 +5,7 @@ # return true if there exist two elements in the BST such that their sum is equal to the given target. # # Example 1: -# Input: +# Input: # 5 # / \ # 3 6 @@ -16,7 +16,7 @@ # # Output: True # Example 2: -# Input: +# Input: # 5 # / \ # 3 6 @@ -51,7 +51,7 @@ def __init__(self, root, forward): def val(self): return self.__cur - + def next(self): while self.__node or self.__s: if self.__node: diff --git a/Python/unique-binary-search-trees-ii.py b/Python/unique-binary-search-trees-ii.py index 1c7fee7ff..1b031830e 100644 --- a/Python/unique-binary-search-trees-ii.py +++ b/Python/unique-binary-search-trees-ii.py @@ -2,10 +2,10 @@ # Space: O(4^n / n^(3/2)) ~= Catalan numbers # # Given n, generate all structurally unique BST's (binary search trees) that store values 1...n. -# +# # For example, # Given n = 3, your program should return all 5 unique BST's shown below. -# +# # 1 3 3 2 1 # \ / / / \ \ # 3 2 1 1 3 2 @@ -19,7 +19,7 @@ def __init__(self, x): self.val = x self.left = None self.right = None - + def __repr__(self): if self: serial = [] @@ -27,21 +27,21 @@ def __repr__(self): while queue: cur = queue[0] - + if cur: serial.append(cur.val) queue.append(cur.left) queue.append(cur.right) else: serial.append("#") - + queue = queue[1:] - + while serial[-1] == "#": serial.pop() - + return repr(serial) - + else: return None @@ -49,7 +49,7 @@ class Solution: # @return a list of tree node def generateTrees(self, n): return self.generateTreesRecu(1, n) - + def generateTreesRecu(self, low, high): result = [] if low > high: diff --git a/Python/unique-binary-search-trees.py b/Python/unique-binary-search-trees.py index e5debe1eb..61d446f51 100644 --- a/Python/unique-binary-search-trees.py +++ b/Python/unique-binary-search-trees.py @@ -2,10 +2,10 @@ # Space: O(1) # # Given n, how many structurally unique BST's (binary search trees) that store values 1...n? -# +# # For example, # Given n = 3, there are a total of 5 unique BST's. -# +# # 1 3 3 2 1 # \ / / / \ \ # 3 2 1 1 3 2 @@ -45,6 +45,6 @@ def numTrees(self, n): count += counts[j] * counts[i - j - 1] counts.append(count) return counts[-1] - + if __name__ == "__main__": print Solution().numTrees(3) diff --git a/Python/unique-morse-code-words.py b/Python/unique-morse-code-words.py index 3b43a876b..8729470dd 100644 --- a/Python/unique-morse-code-words.py +++ b/Python/unique-morse-code-words.py @@ -21,7 +21,7 @@ # Example: # Input: words = ["gin", "zen", "gig", "msg"] # Output: 2 -# Explanation: +# Explanation: # The transformation of each word is: # "gin" -> "--...-." # "zen" -> "--...-." diff --git a/Python/unique-paths-ii.py b/Python/unique-paths-ii.py index ee53e127e..bbe3507b1 100644 --- a/Python/unique-paths-ii.py +++ b/Python/unique-paths-ii.py @@ -2,21 +2,21 @@ # Space: O(m + n) # # Follow up for "Unique Paths": -# +# # Now consider if some obstacles are added to the grids. How many unique paths would there be? -# +# # An obstacle and empty space is marked as 1 and 0 respectively in the grid. -# +# # For example, # There is one obstacle in the middle of a 3x3 grid as illustrated below. -# +# # [ # [0,0,0], # [0,1,0], # [0,0,0] # ] # The total number of unique paths is 2. -# +# # Note: m and n will be at most 100. # @@ -29,7 +29,7 @@ def uniquePathsWithObstacles(self, obstacleGrid): :rtype: int """ m, n = len(obstacleGrid), len(obstacleGrid[0]) - + ways = [0]*n ways[0] = 1 for i in xrange(m): diff --git a/Python/unique-paths.py b/Python/unique-paths.py index c4f234a76..5bfaefa32 100644 --- a/Python/unique-paths.py +++ b/Python/unique-paths.py @@ -2,12 +2,12 @@ # Space: O(m + n) # # A robot is located at the top-left corner of a m x n grid (marked 'Start' in the diagram below). -# +# # The robot can only move either down or right at any point in time. # The robot is trying to reach the bottom-right corner of the grid (marked 'Finish' in the diagram below). -# +# # How many possible unique paths are there? -# +# # Note: m and n will be at most 100. # @@ -17,13 +17,12 @@ def uniquePaths(self, m, n): if m < n: return self.uniquePaths(n, m) ways = [1] * n - + for i in xrange(1, m): for j in xrange(1, n): ways[j] += ways[j - 1] - + return ways[n - 1] if __name__ == "__main__": print Solution().uniquePaths(1, 2) - \ No newline at end of file diff --git a/Python/valid-number.py b/Python/valid-number.py index 75773c0df..047584ba4 100644 --- a/Python/valid-number.py +++ b/Python/valid-number.py @@ -2,7 +2,7 @@ # Space: O(1) # # Validate if a given string is numeric. -# +# # Some examples: # "0" => true # " 0.1 " => true @@ -39,7 +39,7 @@ def isNumber(self, s): [-1, -1, -1, 7, -1, -1], # next states for state 6 [-1, 8, -1, 7, -1, -1], # next states for state 7 [-1, 8, -1, -1, -1, -1]] # next states for state 8 - + state = 0 for char in s: inputType = InputType.INVALID @@ -53,12 +53,12 @@ def isNumber(self, s): inputType = InputType.DOT elif char == 'e' or char == 'E': inputType = InputType.EXPONENT; - + state = transition_table[state][inputType]; - + if state == -1: return False; - + return state == 1 or state == 4 or state == 7 or state == 8 @@ -77,4 +77,4 @@ def isNumber(self, s): print Solution().isNumber("abc") print Solution().isNumber("1 a") print Solution().isNumber("2e10") - + diff --git a/Python/valid-palindrome-ii.py b/Python/valid-palindrome-ii.py index 5e814ca31..3cb4d2d00 100644 --- a/Python/valid-palindrome-ii.py +++ b/Python/valid-palindrome-ii.py @@ -26,7 +26,7 @@ def validPalindrome(s, left, right): return False left, right = left+1, right-1 return True - + left, right = 0, len(s)-1 while left < right: if s[left] != s[right]: diff --git a/Python/valid-palindrome.py b/Python/valid-palindrome.py index c0c562059..fbbcf7f0f 100644 --- a/Python/valid-palindrome.py +++ b/Python/valid-palindrome.py @@ -2,14 +2,14 @@ # Space: O(1) # # Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. -# +# # For example, # "A man, a plan, a canal: Panama" is a palindrome. # "race a car" is not a palindrome. -# +# # Note: # Have you consider that the string might be empty? This is a good question to ask during an interview. -# +# # For the purpose of this problem, we define empty string as valid palindrome. # diff --git a/Python/valid-parentheses.py b/Python/valid-parentheses.py index d62ef17e2..3d9d03781 100644 --- a/Python/valid-parentheses.py +++ b/Python/valid-parentheses.py @@ -3,8 +3,8 @@ # # Given a string containing just the characters '(', ')', '{', '}', '[' and ']', # determine if the input string is valid. -# -# The brackets must close in the correct order, "()" and "()[]{}" +# +# The brackets must close in the correct order, "()" and "()[]{}" # are all valid but "(]" and "([)]" are not. # @@ -18,7 +18,7 @@ def isValid(self, s): elif len(stack) == 0 or lookup[stack.pop()] != parenthese: return False return len(stack) == 0 - + if __name__ == "__main__": print Solution().isValid("()[]{}") print Solution().isValid("()[{]}") \ No newline at end of file diff --git a/Python/valid-sudoku.py b/Python/valid-sudoku.py index 610964276..24f9feb39 100644 --- a/Python/valid-sudoku.py +++ b/Python/valid-sudoku.py @@ -1,10 +1,10 @@ # Time: O(9^2) # Space: O(9) -# Determine if a Sudoku is valid, +# Determine if a Sudoku is valid, # according to: Sudoku Puzzles - The Rules. # -# The Sudoku board could be partially filled, +# The Sudoku board could be partially filled, # where empty cells are filled with the character '.'. # # A partially filled sudoku which is valid. @@ -29,7 +29,7 @@ def isValidSudoku(self, board): for m in xrange(3 * i, 3 * i + 3)]): return False return True - + def isValidList(self, xs): xs = filter(lambda x: x != '.', xs) return len(set(xs)) == len(xs) diff --git a/Python/valid-tic-tac-toe-state.py b/Python/valid-tic-tac-toe-state.py index a8df880cb..7d11d867d 100644 --- a/Python/valid-tic-tac-toe-state.py +++ b/Python/valid-tic-tac-toe-state.py @@ -5,7 +5,7 @@ # if and only if it is possible to reach this board position # during the course of a valid tic-tac-toe game. # -# The board is a 3 x 3 array, and consists of characters " ", "X", +# The board is a 3 x 3 array, and consists of characters " ", "X", # and "O". The " " character represents an empty square. # # Here are the rules of Tic-Tac-Toe: @@ -53,7 +53,7 @@ def win(board, player): return (player == board[1][1] == board[0][0] == board[2][2] or \ player == board[1][1] == board[0][2] == board[2][0]) - + FIRST, SECOND = ('X', 'O') x_count = sum(row.count(FIRST) for row in board) o_count = sum(row.count(SECOND) for row in board) diff --git a/Python/valid-triangle-number.py b/Python/valid-triangle-number.py index 6e1dbbe04..573a27fb2 100644 --- a/Python/valid-triangle-number.py +++ b/Python/valid-triangle-number.py @@ -10,7 +10,7 @@ # Input: [2,2,3,4] # Output: 3 # Explanation: -# Valid combinations are: +# Valid combinations are: # 2,3,4 (using the first 2) # 2,3,4 (using the second 2) # 2,2,3 diff --git a/Python/validate-binary-search-tree.py b/Python/validate-binary-search-tree.py index 8425ef443..d37799d21 100644 --- a/Python/validate-binary-search-tree.py +++ b/Python/validate-binary-search-tree.py @@ -1,10 +1,10 @@ # Time: O(n) # Space: O(1) -# +# # Given a binary tree, determine if it is a valid binary search tree (BST). -# +# # Assume a BST is defined as follows: -# +# # The left subtree of a node contains only nodes with keys less than the node's key. # The right subtree of a node contains only nodes with keys greater than the node's key. # Both the left and right subtrees must also be binary search trees. @@ -33,7 +33,7 @@ def isValidBST(self, root): node = cur.left while node.right and node.right != cur: node = node.right - + if node.right is None: node.right = cur cur = cur.left @@ -43,7 +43,7 @@ def isValidBST(self, root): node.right = None prev = cur cur = cur.right - + return True @@ -54,16 +54,16 @@ class Solution2: # @return a boolean def isValidBST(self, root): return self.isValidBSTRecu(root, float("-inf"), float("inf")) - + def isValidBSTRecu(self, root, low, high): if root is None: return True - + return low < root.val and root.val < high \ and self.isValidBSTRecu(root.left, low, root.val) \ and self.isValidBSTRecu(root.right, root.val, high) - + if __name__ == "__main__": root = TreeNode(2) root.left = TreeNode(1) diff --git a/Python/verify-preorder-serialization-of-a-binary-tree.py b/Python/verify-preorder-serialization-of-a-binary-tree.py index ee7d23f75..c084b07ca 100644 --- a/Python/verify-preorder-serialization-of-a-binary-tree.py +++ b/Python/verify-preorder-serialization-of-a-binary-tree.py @@ -2,7 +2,7 @@ # Space: O(1) # One way to serialize a binary tree is to use pre-oder traversal. -# When we encounter a non-null node, we record the node's value. +# When we encounter a non-null node, we record the node's value. # If it is a null node, we record using a sentinel value such as #. # # _9_ @@ -15,8 +15,8 @@ # For example, the above binary tree can be serialized to the string # "9,3,4,#,#,1,#,#,2,#,6,#,#", where # represents a null node. # -# Given a string of comma separated values, verify whether it is a -# correct preorder traversal serialization of a binary tree. +# Given a string of comma separated values, verify whether it is a +# correct preorder traversal serialization of a binary tree. # Find an algorithm without reconstructing the tree. # # Each comma separated value in the string must be either an integer diff --git a/Python/walls-and-gates.py b/Python/walls-and-gates.py index c3113ffd2..7d287cfcc 100644 --- a/Python/walls-and-gates.py +++ b/Python/walls-and-gates.py @@ -18,4 +18,4 @@ def wallsAndGates(self, rooms): rooms[I][J] == INF: rooms[I][J] = rooms[i][j] + 1 q.append((I, J)) - + diff --git a/Python/water-and-jug-problem.py b/Python/water-and-jug-problem.py index b6dcfe954..df6f3601d 100644 --- a/Python/water-and-jug-problem.py +++ b/Python/water-and-jug-problem.py @@ -22,7 +22,7 @@ # Input: x = 2, y = 6, z = 5 # Output: False -# Bézout's identity (also called Bézout's lemma) +# Bézout's identity (also called Bézout's lemma) class Solution(object): def canMeasureWater(self, x, y, z): """ @@ -35,7 +35,7 @@ def gcd(a, b): while b: a, b = b, a%b return a - + # The problem is to solve: # - check z <= x + y # - check if there is any (a, b) integers s.t. ax + by = z diff --git a/Python/wiggle-sort-ii.py b/Python/wiggle-sort-ii.py index 15b625f07..58efd5258 100644 --- a/Python/wiggle-sort-ii.py +++ b/Python/wiggle-sort-ii.py @@ -4,7 +4,7 @@ # Given an unsorted array nums, reorder it such that nums[0] < nums[1] > nums[2] < nums[3].... # # Example: -# (1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6]. +# (1) Given nums = [1, 5, 1, 1, 6, 4], one possible answer is [1, 4, 1, 5, 1, 6]. # (2) Given nums = [1, 3, 2, 2, 3, 1], one possible answer is [2, 3, 1, 3, 1, 2]. # # Note: @@ -45,7 +45,7 @@ def findKthLargest(nums, k): right = new_pivot_idx - 1 else: # new_pivot_idx < k - 1. left = new_pivot_idx + 1 - + def partitionAroundPivot(left, right, pivot_idx, nums): pivot_value = nums[pivot_idx] new_pivot_idx = left diff --git a/Python/wiggle-subsequence.py b/Python/wiggle-subsequence.py index 3336a9338..abb8e6148 100644 --- a/Python/wiggle-subsequence.py +++ b/Python/wiggle-subsequence.py @@ -9,7 +9,7 @@ # is trivially a wiggle sequence. # # For example, [1,7,4,9,2,5] is a wiggle sequence because -# the differences (6,-3,5,-7,3) are alternately positive +# the differences (6,-3,5,-7,3) are alternately positive # and negative. In contrast, [1,4,7,2,5] and [1,7,4,5,5] are # not wiggle sequences, the first because its first two differences # are positive and the second because its last difference is zero. @@ -17,7 +17,7 @@ # Given a sequence of integers, return the length of # the longest subsequence that is a wiggle sequence. # A subsequence is obtained by deleting some number of elements -# (eventually, also zero) from the original sequence, leaving +# (eventually, also zero) from the original sequence, leaving # the remaining elements in their original order. # # Examples: diff --git a/Python/wildcard-matching.py b/Python/wildcard-matching.py index 7965368aa..f98db3e20 100644 --- a/Python/wildcard-matching.py +++ b/Python/wildcard-matching.py @@ -2,15 +2,15 @@ # Space: O(1) # # Implement wildcard pattern matching with support for '?' and '*'. -# +# # '?' Matches any single character. # '*' Matches any sequence of characters (including the empty sequence). -# +# # The matching should cover the entire input string (not partial). -# +# # The function prototype should be: # bool isMatch(const char *s, const char *p) -# +# # Some examples: # isMatch("aa","a") -> false # isMatch("aa","aa") -> true @@ -42,12 +42,12 @@ def isMatch(self, s, p): p_ptr = last_p_ptr else: return False - + while p_ptr < len(p) and p[p_ptr] == '*': p_ptr += 1 - + return p_ptr == len(p) - + # dp with rolling window # Time: O(m * n) # Space: O(m + n) @@ -68,7 +68,7 @@ def isMatch(self, s, p): result[i % k][j] = result[(i-1) % k][j-1] and (s[i-1] == p[j-1] or p[j-1] == '?') else: result[i % k][j] = result[i % k][j-1] or result[(i-1) % k][j] - + return result[len(s) % k][len(p)] # dp @@ -90,7 +90,7 @@ def isMatch(self, s, p): result[i][j] = result[i-1][j-1] and (s[i-1] == p[j-1] or p[j-1] == '?') else: result[i][j] = result[i][j-1] or result[i-1][j] - + return result[len(s)][len(p)] @@ -100,7 +100,7 @@ class Solution4: def isMatch(self, s, p): if not p or not s: return not s and not p - + if p[0] != '*': if p[0] == s[0] or p[0] == '?': return self.isMatch(s[1:], p[1:]) diff --git a/Python/word-break-ii.py b/Python/word-break-ii.py index 7dcb65c6d..e23c7ec25 100644 --- a/Python/word-break-ii.py +++ b/Python/word-break-ii.py @@ -1,16 +1,16 @@ -# Time: O(n * l^2 + n * r), l is the max length of the words, +# Time: O(n * l^2 + n * r), l is the max length of the words, # r is the number of the results. # Space: O(n^2) # -# Given a string s and a dictionary of words dict, +# Given a string s and a dictionary of words dict, # add spaces in s to construct a sentence where each word is a valid dictionary word. -# +# # Return all such possible sentences. -# +# # For example, given # s = "catsanddog", # dict = ["cat", "cats", "and", "sand", "dog"]. -# +# # A solution is ["cats and dog", "cat sand dog"]. # @@ -40,7 +40,7 @@ def wordBreak(self, s, wordDict): if can_break[-1]: self.wordBreakHelper(s, valid, 0, [], result) return result - + def wordBreakHelper(self, s, valid, start, path, result): if start == len(s): result.append(" ".join(path)) diff --git a/Python/word-break.py b/Python/word-break.py index e64e5631e..5955525de 100644 --- a/Python/word-break.py +++ b/Python/word-break.py @@ -1,13 +1,13 @@ # Time: O(n * l^2) # Space: O(n) -# Given a string s and a dictionary of words dict, +# Given a string s and a dictionary of words dict, # determine if s can be segmented into a space-separated sequence of one or more dictionary words. -# +# # For example, given # s = "leetcode", # dict = ["leet", "code"]. -# +# # Return true because "leetcode" can be segmented as "leet code". class Solution(object): @@ -33,6 +33,6 @@ def wordBreak(self, s, wordDict): return can_break[-1] - + if __name__ == "__main__": print Solution().wordBreak("leetcode", ["leet", "code"]) diff --git a/Python/word-ladder-ii.py b/Python/word-ladder-ii.py index b7ffb1180..7edc61b1c 100644 --- a/Python/word-ladder-ii.py +++ b/Python/word-ladder-ii.py @@ -2,11 +2,11 @@ # Space: O(d) # # Given two words (start and end), and a dictionary, find all shortest transformation sequence(s) from start to end, such that: -# +# # Only one letter can be changed at a time # Each intermediate word must exist in the dictionary # For example, -# +# # Given: # start = "hit" # end = "cog" @@ -30,13 +30,13 @@ class Solution: def findLadders(self, start, end, dict): dict.add(start) dict.add(end) - - result, cur, visited, found, trace = [], [start], set([start]), False, {word: [] for word in dict} + + result, cur, visited, found, trace = [], [start], set([start]), False, {word: [] for word in dict} while cur and not found: for word in cur: visited.add(word) - + next = set() for word in cur: for i in xrange(len(word)): @@ -48,18 +48,18 @@ def findLadders(self, start, end, dict): next.add(candidate) trace[candidate].append(word) cur = next - + if found: self.backtrack(result, trace, [], end) - + return result - + def backtrack(self, result, trace, path, word): if not trace[word]: result.append([word] + path) else: for prev in trace[word]: self.backtrack(result, trace, [word] + path, prev) - + if __name__ == "__main__": print Solution().findLadders("hit", "cog", set(["hot","dot","dog","lot","log"])) diff --git a/Python/word-ladder.py b/Python/word-ladder.py index 3a97d25a8..ff2b0cb70 100644 --- a/Python/word-ladder.py +++ b/Python/word-ladder.py @@ -2,18 +2,18 @@ # Space: O(d) # # Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that: -# +# # Only one letter can be changed at a time # Each intermediate word must exist in the dictionary # For example, -# +# # Given: # start = "hit" # end = "cog" # dict = ["hot","dot","dog","lot","log"] # As one shortest transformation is "hit" -> "hot" -> "dot" -> "dog" -> "cog", # return its length 5. -# +# # Note: # Return 0 if there is no such transformation sequence. # All words have the same length. diff --git a/Python/word-pattern-ii.py b/Python/word-pattern-ii.py index 2d12cd67c..774795858 100644 --- a/Python/word-pattern-ii.py +++ b/Python/word-pattern-ii.py @@ -1,7 +1,7 @@ # Time: O(n * C(n - 1, c - 1)), n is length of str, c is unique count of pattern, # there are H(n - c, c - 1) = C(n - 1, c - 1) possible splits of string, # and each one costs O(n) to check if it matches the word pattern. -# Space: O(n + c) +# Space: O(n + c) class Solution(object): def wordPatternMatch(self, pattern, str): diff --git a/Python/word-pattern.py b/Python/word-pattern.py index 7bace4ebd..ef0db71b9 100644 --- a/Python/word-pattern.py +++ b/Python/word-pattern.py @@ -32,12 +32,12 @@ def wordPattern(self, pattern, str): if w not in w2p and p not in p2w: # Build mapping. Space: O(c) w2p[w] = p - p2w[p] = w + p2w[p] = w elif w not in w2p or w2p[w] != p: # Contradict mapping. return False return True - + def wordCount(self, str): cnt = 1 if str else 0 for c in str: @@ -75,7 +75,7 @@ def wordPattern(self, pattern, str): if w not in w2p and p not in p2w: # Build mapping. Space: O(c) w2p[w] = p - p2w[p] = w + p2w[p] = w elif w not in w2p or w2p[w] != p: # Contradict mapping. return False diff --git a/Python/word-search-ii.py b/Python/word-search-ii.py index 4c6acdaac..c317e6b2d 100644 --- a/Python/word-search-ii.py +++ b/Python/word-search-ii.py @@ -3,7 +3,7 @@ # # Given a 2D board and a list of words from the dictionary, find all words in the board. # -# Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells +# Each word must be constructed from letters of sequentially adjacent cell, where "adjacent" cells # are those horizontally or vertically neighboring. The same letter cell may not be used more than once in a word. # # For example, @@ -48,30 +48,30 @@ def findWords(self, board, words): trie = TrieNode() for word in words: trie.insert(word) - + for i in xrange(len(board)): for j in xrange(len(board[0])): if self.findWordsRecu(board, trie, 0, i, j, visited, [], result): return True - + return result.keys() - + def findWordsRecu(self, board, trie, cur, i, j, visited, cur_word, result): if not trie or i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or visited[i][j]: return - + if board[i][j] not in trie.leaves: return - + cur_word.append(board[i][j]) next_node = trie.leaves[board[i][j]] if next_node.is_string: result["".join(cur_word)] = True - + visited[i][j] = True self.findWordsRecu(board, next_node, cur + 1, i + 1, j, visited, cur_word, result) self.findWordsRecu(board, next_node, cur + 1, i - 1, j, visited, cur_word, result) self.findWordsRecu(board, next_node, cur + 1, i, j + 1, visited, cur_word, result) - self.findWordsRecu(board, next_node, cur + 1, i, j - 1, visited, cur_word, result) + self.findWordsRecu(board, next_node, cur + 1, i, j - 1, visited, cur_word, result) visited[i][j] = False cur_word.pop() diff --git a/Python/word-search.py b/Python/word-search.py index 3166d0b00..719540086 100644 --- a/Python/word-search.py +++ b/Python/word-search.py @@ -2,14 +2,14 @@ # Space: O(l) # # Given a 2D board and a word, find if the word exists in the grid. -# -# The word can be constructed from letters of sequentially adjacent cell, -# where "adjacent" cells are those horizontally or vertically neighboring. +# +# The word can be constructed from letters of sequentially adjacent cell, +# where "adjacent" cells are those horizontally or vertically neighboring. # The same letter cell may not be used more than once. -# +# # For example, # Given board = -# +# # [ # "ABCE", # "SFCS", @@ -26,30 +26,30 @@ class Solution: # @return a boolean def exist(self, board, word): visited = [[False for j in xrange(len(board[0]))] for i in xrange(len(board))] - + for i in xrange(len(board)): for j in xrange(len(board[0])): if self.existRecu(board, word, 0, i, j, visited): return True - + return False - + def existRecu(self, board, word, cur, i, j, visited): if cur == len(word): return True - + if i < 0 or i >= len(board) or j < 0 or j >= len(board[0]) or visited[i][j] or board[i][j] != word[cur]: return False - + visited[i][j] = True result = self.existRecu(board, word, cur + 1, i + 1, j, visited) or\ self.existRecu(board, word, cur + 1, i - 1, j, visited) or\ self.existRecu(board, word, cur + 1, i, j + 1, visited) or\ - self.existRecu(board, word, cur + 1, i, j - 1, visited) + self.existRecu(board, word, cur + 1, i, j - 1, visited) visited[i][j] = False - + return result - + if __name__ == "__main__": board = [ "ABCE", diff --git a/Python/word-squares.py b/Python/word-squares.py index 59d69b9e6..21d6bbf43 100644 --- a/Python/word-squares.py +++ b/Python/word-squares.py @@ -34,11 +34,11 @@ def wordSquares(self, words): curr.pop() return result - + def wordSquaresHelper(self, words, trie, curr, result): if len(curr) >= len(words[0]): return result.append(list(curr)) - + node = trie for s in curr: node = node.children[ord(s[len(curr)]) - ord('a')] diff --git a/Python/zigzag-conversion.py b/Python/zigzag-conversion.py index f5381a1b5..5a1d04cee 100644 --- a/Python/zigzag-conversion.py +++ b/Python/zigzag-conversion.py @@ -1,15 +1,15 @@ # Time: O(n) # Space: O(1) # -# The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: +# The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: # (you may want to display this pattern in a fixed font for better legibility) -# +# # P A H N # A P L S I I G # Y I R # And then read line by line: "PAHNAPLSIIGYIR" # Write the code that will take a string and make this conversion given a number of rows: -# +# # string convert(string text, int nRows); # convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR". #