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[yuseok89] WEEK 04 Solutions #2739
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. bfs 방식으로 풀어주셨네요. dict 이용해서 필요한 공간만 사용하는 것도 신경쓰셔서 푸신 거 같네요.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 큐를 이용해 현재 포함된 합에 도달한 최소 동전 수를 업데이트한다. 각 합은 한 번만 방문하므로 전체 시간은 금액 범위와 동전 종류에 비례한다. 개선 제안: 현재 구현이 적절해 보입니다. |
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| # TC: O(amount * len(coins)) | ||
| # SC: O(amount) | ||
| class Solution: | ||
| def coinChange(self, coins: List[int], amount: int) -> int: | ||
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| from collections import deque | ||
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| v = {0: 0} | ||
| q = deque([0]) | ||
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| while q and amount not in v: | ||
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| cur = q.popleft() | ||
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| for coin in coins: | ||
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| if cur + coin > amount or cur + coin in v: | ||
| continue | ||
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| v[cur + coin] = v[cur] + 1 | ||
| q.append(cur + coin) | ||
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| return -1 if amount not in v else v[amount] | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 저는 while loop 안에 복잡한 연산 들어가는거 별로라 최대한 간단하게 했는데, 이렇게 하시면 얼리 리턴이 가능해서 좋네요!
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 범위를 반으로 나누며 조건에 따라 최소값의 위치를 좁혀간다. 개선 제안: 현재 구현이 적절해 보입니다. |
| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,20 @@ | ||
| # TC: O(logN) | ||
| # SC: O(1) | ||
| class Solution: | ||
| def findMin(self, nums: List[int]) -> int: | ||
| low = 0 | ||
| high = len(nums) - 1 | ||
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| while low < high - 1: | ||
| mid = (low + high) // 2 | ||
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| if nums[low] < nums[mid] < nums[high]: | ||
| return nums[low] | ||
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| if nums[low] > nums[mid]: | ||
| high = mid | ||
| else: | ||
| low = mid | ||
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yuseok89 marked this conversation as resolved.
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| return min(nums[low], nums[high]) | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 아주 개인적인 생각인데, rec을 쓰지 않고 maxDepth 자체를 재귀로 하면 좀 코드가 간단해지지 않을까요?
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 그러네요. 문제 풀 당시에는 바로 생각을 못했었네요.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 각 노드를 한 번씩 방문하고 호출 스택의 깊이는 트리의 높이에 비례한다. 개선 제안: 현재 구현이 적절해 보입니다. |
| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,19 @@ | ||
| # TC: O(N) | ||
| # SC: O(H) | ||
| # 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 maxDepth(self, root: Optional[TreeNode]) -> int: | ||
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| def rec(node, height): | ||
| if not node: | ||
| return height | ||
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| return max(rec(node.left, height + 1), rec(node.right, height + 1)) | ||
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| return rec(root, 0) | ||
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 이거 2개 비교하는 부분 끝나고 리스트 두개중에 한개 순회 다 끝난 상태에서는 나머지 남은 리스트는 그냥 뒤에 통째로 붙혀주는게 더 좋더라구요!
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 의견 감사합니다.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 두 리스트를 동시에 순회하며 더 작은 노드를 연결한다. 개선 제안: 현재 구현이 적절해 보입니다. |
| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,26 @@ | ||
| # TC: O(N+M) | ||
| # SC: O(1) | ||
| # 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]: | ||
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| ret = ListNode() | ||
| cur = ret | ||
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. ret은 무슨 약자일까요?
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Author
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. return value 약자로 썼는데, |
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| while list1 and list2: | ||
| if list1.val < list2.val: | ||
| cur.next = list1 | ||
| list1 = list1.next | ||
| else: | ||
| cur.next = list2 | ||
| list2 = list2.next | ||
| cur = cur.next | ||
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| cur.next = list1 if list1 else list2 | ||
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| return ret.next | ||
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Contributor
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 이거 회원님 답안은 3000ms 가량 걸리는데
이 두가지만 추가해도 leetcode 실행속도상 0ms까지 도달할수 있어요!
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 백트래킹에서 탐색공간을 프루닝하는건 좋은 공부가 될것 같습니다!
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 리뷰 감사합니다.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 맞습니다! 빈도 기반으로 뒤집었는데, 해당 문제를 어떻게 하면 조금이라도 빨리 풀수 있을까에만 초점을 맞춘 코드이긴 합니다!
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 입력 글자 빈도 검사를 먼저 수행해 실패를 조기에 핸들링한다. DFS로 인접 칸을 탐색한다. 개선 제안: 현재 구현이 적절해 보입니다. |
| Original file line number | Diff line number | Diff line change |
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| @@ -0,0 +1,37 @@ | ||
| # TC: O(N*M*4^L) | ||
| # SC: O(L) | ||
| class Solution: | ||
| def exist(self, board: List[List[str]], word: str) -> bool: | ||
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| n = len(board) | ||
| m = len(board[0]) | ||
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| board_counter = Counter(board[r][c] for r in range(n) for c in range(m)) | ||
| word_counter = Counter(word) | ||
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| for c, cnt in word_counter.items(): | ||
| if board_counter[c] < cnt: | ||
| return False | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 빈도 이용해서 최적화하신 거 좋은 거 같습니다. 이걸 rec 함수에서도 사용하면 좀 더 최적화가 가능할 거 같네요.
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. rec 함수에서 어떻게 활용할 수 있을까요?
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 제가 잘못 생각헀네요. 초기에 한 번 사용하면 이후 과정에선 불필요하네요! |
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| def rec(row, col, idx): | ||
| if idx == len(word): | ||
| return True | ||
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| if row < 0 or row >= n or col < 0 or col >= m or board[row][col] != word[idx]: | ||
| return False | ||
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| board[row][col] = None | ||
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| ret = rec(row + 1, col, idx + 1) or rec(row - 1, col, idx + 1) or rec(row, col + 1, idx + 1) or rec(row, col - 1, idx + 1) | ||
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| board[row][col] = word[idx] | ||
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There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. None으로 할당했다가 word[idx]로 복원하는거 좋은 거 같습니다! |
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| return ret | ||
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| for row in range(0, n): | ||
| for col in range(0, m): | ||
| if board[row][col] == word[0] and rec(row, col, 0): | ||
| return True | ||
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| return False | ||
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저는 코인이 양수값만 있어서 DP만 했었는데 BFS로 해도 충분히 가능하겠네요!