[bhyun-kim, 김병현] Solution of Week 3 Assignments

climbing-stairs/bhyun-kim.py

@@ -0,0 +1,39 @@
+"""
+70. Climbing Stairs
+https://leetcode.com/problems/climbing-stairs/description/
+
+Solution:
+    This is a combinatorial problem that allows duplicates.
+    We can use the formula for combinations to solve this problem.
+    Let's suppose that n is the number of steps, and k is the number of 2-steps.
+    Then, the number of ways to climb the stairs is n!/((n-2k)!k!).
+    We can iterate through all possible values of k, and calculate the number of ways to climb the stairs.
+
+    1. Create a dictionary to store the factorials of numbers from 0 to n.
+    2. Calculate the factorials of numbers from 0 to n.
+    3. Iterate through all possible values of k from 0 to n//2.
+
+
+Time complexity: O(n)
+Space complexity: O(n)   
+"""
+
+from collections import defaultdict
+
+
+class Solution:
+    def climbStairs(self, n: int) -> int:
+        factorials = defaultdict(int)
+        factorials[0] = 1
+
+        fact = 1
+        for i in range(1, n + 1):
+            fact = fact * i
+            factorials[i] = fact
+
+        output = 0
+
+        for i in range(n // 2 + 1):
+            num_ways = factorials[n - i] / factorials[n - (i * 2)] / factorials[i]
+            output += num_ways
+        return int(output)

invert-binary-tree/bhyun-kim.py

@@ -31,7 +31,9 @@ def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]:
         has_right = hasattr(root, "right")
 
         if has_left and has_right:
-            root.left, root.right = self.invertTree(root.right), self.invertTree(root.left)
+            root.left, root.right = self.invertTree(root.right), self.invertTree(
+                root.left
+            )
         elif has_left:
             root.left, root.right = None, self.invertTree(root.left)
         elif has_right:

maximum-depth-of-binary-tree/bhyun-kim.py

@@ -0,0 +1,41 @@
+"""
+104. Maximum Depth of Binary Tree
+https://leetcode.com/problems/maximum-depth-of-binary-tree/
+
+Solution:
+    This is a recursive problem that requires traversing the binary tree.
+    We can use a recursive function to traverse the binary tree and calculate the depth.
+    The depth of the binary tree is the maximum of the depth of the left and right subtrees.
+    We can calculate the depth of the left and right subtrees recursively.
+    The base case is when the root is None, in which case the depth is 0.
+    
+    1. Calculate the depth of the left subtree.
+    2. Calculate the depth of the right subtree.
+    3. Return the maximum of the depth of the left and right subtrees plus 1.
+
+Time complexity: O(n)
+Space complexity: O(n)
+"""
+
+
+from typing import Optional
+
+
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def maxDepth(self, root: Optional[TreeNode]) -> int:
+        max_dep_left = 0
+        max_dep_right = 0
+
+        if hasattr(root, "left"):
+            max_dep_left = 1 + self.maxDepth(root.left)
+        if hasattr(root, "right"):
+            max_dep_right = 1 + self.maxDepth(root.right)
+
+        return max(max_dep_left, max_dep_right)

meeting-rooms/bhyun-kim.py

@@ -0,0 +1,55 @@
+"""
+252. Meeting Rooms
+https://leetcode.com/problems/meeting-rooms/
+
+Solution:
+    This is a sorting problem that requires comparing intervals.
+    We can sort the intervals by the start time.
+    Then, we can iterate through the sorted intervals and check if the end time of the current interval is greater than the start time of the next interval.
+    If it is, then we return False.
+    Otherwise, we return True.
+    
+    1. Sort the intervals by the start time.
+    2. Iterate through the sorted intervals.
+    3. Check if the end time of the current interval is greater than the start time of the next interval.
+
+
+Time complexity: O(nlogn)
+Space complexity: O(1)
+
+Discarded solution:
+    This solution uses a hash table to store the time table of the intervals.
+    We iterate through the intervals and check if there is any overlap in the time table.
+
+    1. Create a hash table to store the time table of the intervals.
+    2. Iterate through the intervals.
+    3. Check if there is any overlap in the time table.
+    4. Return False if there is an overlap, otherwise return True.
+
+Time complexity: O(n^2)
+Space complexity: O(n)
+
+class Solution:
+    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
+
+        time_table = defaultdict(int)
+        for itrv in intervals: 
+            for i in range(itrv[0], itrv[1]):
+                if time_table[i] == 0:
+                    time_table[i] += 1 
+                else:
+                    return False 
+        return True
+"""
+
+
+from typing import List
+
+
+class Solution:
+    def canAttendMeetings(self, intervals: List[List[int]]) -> bool:
+        intervals.sort()
+        for i in range(len(intervals) - 1):
+            if intervals[i][1] > intervals[i + 1][0]:
+                return False
+        return True

same-tree/bhyun-kim.py

@@ -0,0 +1,43 @@
+"""
+100. Same Tree
+https://leetcode.com/problems/same-tree/description/
+
+Solution:
+    This is a recursive problem that requires comparing two binary trees.
+    We can use a recursive function to compare the values of the nodes in the two binary trees.
+    If the values of the nodes are equal, we can compare the left and right subtrees recursively.
+    The base case is when both nodes are None, in which case we return True.
+    
+    1. Check if the values of the nodes are equal.
+    2. Compare the left and right subtrees recursively.
+    3. Return True if the values of the nodes are equal and the left and right subtrees are equal, otherwise return False.
+"""
+
+
+from typing import Optional
+
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
+        if p == q == None:
+            return True
+
+        if hasattr(p, "val") and hasattr(q, "val"):
+            if p.val != q.val:
+                return False
+
+            left_same = self.isSameTree(p.left, q.left)
+            right_same = self.isSameTree(p.right, q.right)
+
+            return left_same == right_same == True
+
+        else:
+            return False

subtree-of-another-tree/bhyun-kim.py

@@ -0,0 +1,75 @@
+"""
+572. Subtree of Another Tree
+https://leetcode.com/problems/subtree-of-another-tree/description/
+
+Solution:
+    This solution uses a depth-first search to find the subtree in the tree.
+    We can use a recursive function to traverse the tree and compare the nodes.
+    If the given node is identical to the subtree, we return True.
+    Otherwise, we go to the left and right subtrees recursively.
+    In this solution we use two helper functions: depth_first_search and is_identical.
+
+    Depth-first search:
+    1. Check if the root is None.
+    2. Check if the root is identical to the subtree.
+    3. Go to the left and right subtrees recursively.
+
+    Is identical:
+    1. Check if both nodes are None.
+    2. Check if one of the nodes is None.
+    3. Check if the values of the nodes are equal.
+    4. Compare the left and right subtrees recursively.
+
+
+Complexity analysis:
+    Time complexity: O(n*m)
+        Where n is the number of nodes in the tree and m is the number of nodes in the subtree.
+        The depth-first search function has a time complexity of O(n).
+        The is_identical function has a time complexity of O(m).
+        Therefore, the overall time complexity is O(n*m).
+
+    Space complexity: O(n)
+        Where n is the number of nodes in the tree.
+        The space complexity is O(n) because of the recursive calls to the depth_first_search function.
+"""
+
+
+from typing import Optional
+
+
+# Definition for a binary tree node.
+class TreeNode:
+    def __init__(self, val=0, left=None, right=None):
+        self.val = val
+        self.left = left
+        self.right = right
+
+
+class Solution:
+    def isSubtree(self, root: Optional[TreeNode], subRoot: Optional[TreeNode]) -> bool:
+        def depth_first_search(root):
+            if root is None:
+                return False
+
+            if is_identical(root, subRoot):
+                return True
+
+            return depth_first_search(root.left) or depth_first_search(root.right)
+
+        def is_identical(root1, root2):
+            if root1 is None and root2 is None:
+                return True
+            if root1 is not None and root2 is None:
+                return False
+            if root1 is None and root2 is not None:
+                return False
+            if root1.val == root2.val:
+                return (
+                    is_identical(root1.left, root2.left)
+                    == is_identical(root1.right, root2.right)
+                    == True
+                )
+            else:
+                return False
+
+        return depth_first_search(root)