Merge pull request #819 from Chaedie/main

[Chaedie] Week 4

Author: Chaedie

coin-change/Chaedie.py

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+"""
+직접 풀지 못해 알고달레 풀이를 참고했습니다. https://www.algodale.com/problems/coin-change/
+
+Solution:
+    1) BFS를 통해 모든 동전을 한번씩 넣어보며 amount와 같아지면 return
+
+(c: coins의 종류 갯수, a: amount)
+Time: O(ca)
+Space: O(a)
+"""
+
+
+class Solution:
+    def coinChange(self, coins: List[int], amount: int) -> int:
+        q = deque([(0, 0)])  # (동전 갯수, 누적 금액)
+        visited = set()
+
+        while q:
+            count, total = q.popleft()
+            if total == amount:
+                return count
+            if total in visited:
+                continue
+            visited.add(total)
+            for coin in coins:
+                if total + coin <= amount:
+                    q.append((count + 1, total + coin))
+        return -1

merge-two-sorted-lists/Chaedie.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def mergeTwoLists(
+        self, list1: Optional[ListNode], list2: Optional[ListNode]
+    ) -> Optional[ListNode]:
+        """
+        Solution:
+            1) 리스트1 리스트2가 null 이 아닌 동안 list1, list2를 차례대로 줄세운다.
+            2) list1이 남으면 node.next = list1로 남은 리스트를 연결한다.
+            3) list2가 남으면 list2 를 연결한다.
+
+        Time: O(n)
+        Space: O(1)
+        """
+        dummy = ListNode()
+        node = dummy
+
+        while list1 and list2:
+            if list1.val < list2.val:
+                node.next = list1
+                list1 = list1.next
+            else:
+                node.next = list2
+                list2 = list2.next
+            node = node.next
+
+        if list1:
+            node.next = list1
+        elif list2:
+            node.next = list2
+
+        return dummy.next

missing-number/Chaedie.py

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+"""
+Solution: 
+    1) 배열 정렬
+    2) 0부터 for 문을 돌리는데 index 와 값이 다르면 return index
+    3) 끝까지 일치한다면 return 배열의 크기 
+Time: O(nlogn) = O(nlogn) + O(n)
+Space: O(1)
+"""
+
+
+class Solution:
+    def missingNumber(self, nums: List[int]) -> int:
+        nums.sort()
+        for i in range(len(nums)):
+            if i != nums[i]:
+                return i
+        return len(nums)

palindromic-substrings/Chaedie.py

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+"""
+Solution: 
+    1) 자신을 기준으로 l,r 포인터로 늘려주면서 같은 문자이면 palindrome
+        이를 홀수, 짝수 글자에 대해 2번 진행해주면된다.
+Time: O(n^2) = O(n) * O(n/2 * 2)
+Space: O(1)
+    
+"""
+
+
+class Solution:
+    def countSubstrings(self, s: str) -> int:
+        result = 0
+        for i in range(len(s)):
+            l, r = i, i
+            while l >= 0 and r < len(s):
+                if s[l] != s[r]:
+                    break
+                l -= 1
+                r += 1
+                result += 1
+
+            l, r = i, i + 1
+            while l >= 0 and r < len(s):
+                if s[l] != s[r]:
+                    break
+                l -= 1
+                r += 1
+                result += 1
+        return result

word-search/Chaedie.py

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+"""
+Solution: 
+    1) board의 상하좌우를 탐색하되 아래 조건을 base case로 걸러준다.
+    1.1) index 가 word의 길이이면 결과값 판단
+    1.2) out of bounds 판단
+    1.3) index를 통해 현재 글자와 board의 글자의 일치 판단
+    1.4) 방문 여부 판단
+    2) board를 돌면서 backtrack 이 True 인 케이스가 있으면 return True
+
+m = row_len
+n = col_len
+L = 단어 길이
+Time: O(m n 4^L)
+Space: O(mn + L^2) = visit set O(mn) + 호출 스택 및 cur_word O(L^2)
+"""
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+        visit = set()
+
+        def backtrack(r, c, index, cur_word):
+            if index == len(word):
+                return word == cur_word
+            if r < 0 or c < 0 or r == ROWS or c == COLS:
+                return False
+            if word[index] != board[r][c]:
+                return False
+            if (r, c) in visit:
+                return False
+
+            visit.add((r, c))
+            condition = (
+                backtrack(r + 1, c, index + 1, cur_word + board[r][c])
+                or backtrack(r - 1, c, index + 1, cur_word + board[r][c])
+                or backtrack(r, c + 1, index + 1, cur_word + board[r][c])
+                or backtrack(r, c - 1, index + 1, cur_word + board[r][c])
+            )
+            visit.remove((r, c))
+            return condition
+
+        for i in range(ROWS):
+            for j in range(COLS):
+                if backtrack(i, j, 0, ""):
+                    return True
+        return False
+
+
+"""
+Solution: 
+    공간 복잡도 낭비를 줄이기 위해 cur_word 제거
+Time: O(m n 4^L)
+Space: O(mn + L)
+"""
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+        visit = set()
+
+        def backtrack(r, c, index):
+            if index == len(word):
+                return True
+            if r < 0 or c < 0 or r == ROWS or c == COLS:
+                return False
+            if word[index] != board[r][c]:
+                return False
+            if (r, c) in visit:
+                return False
+
+            visit.add((r, c))
+            condition = (
+                backtrack(r + 1, c, index + 1)
+                or backtrack(r - 1, c, index + 1)
+                or backtrack(r, c + 1, index + 1)
+                or backtrack(r, c - 1, index + 1)
+            )
+            visit.remove((r, c))
+            return condition
+
+        for i in range(ROWS):
+            for j in range(COLS):
+                if backtrack(i, j, 0):
+                    return True
+        return False
+
+
+"""
+Solution: 
+    공간 복잡도를 줄이기 위해 visit set 제거 
+    -> board[r][c]에 빈문자열을 잠깐 추가하는것으로 대체
+Time: O(m n 4^L)
+Space: O(L)
+"""
+
+
+class Solution:
+    def exist(self, board: List[List[str]], word: str) -> bool:
+        ROWS, COLS = len(board), len(board[0])
+
+        def backtrack(r, c, index):
+            if index == len(word):
+                return True
+            if r < 0 or c < 0 or r == ROWS or c == COLS:
+                return False
+            if word[index] != board[r][c]:
+                return False
+
+            temp = board[r][c]
+            board[r][c] = ""
+            condition = (
+                backtrack(r + 1, c, index + 1)
+                or backtrack(r - 1, c, index + 1)
+                or backtrack(r, c + 1, index + 1)
+                or backtrack(r, c - 1, index + 1)
+            )
+            board[r][c] = temp
+            return condition
+
+        for i in range(ROWS):
+            for j in range(COLS):
+                if backtrack(i, j, 0):
+                    return True
+        return False