-
-
Notifications
You must be signed in to change notification settings - Fork 361
[yuseok89] WEEK 04 Solutions #2739
New issue
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
base: main
Are you sure you want to change the base?
Changes from all commits
4836151
cdc00da
cc0a988
c894055
4e4e412
32cafb9
49ecb59
a6fcc33
38a1f67
8bd5a32
05685ed
cd23e34
120df1f
1d4ff54
23142a7
0cb14ca
9465142
8b4a01f
4b09054
da3aad9
5c4d5fb
03f3f1a
d8d5ea4
1f003e2
a022d10
b8248c4
51c1012
65431c0
e514ab2
194db1c
419dfc5
af586d8
e2f0e02
41107c5
37ecdd8
fc56573
File filter
Filter by extension
Conversations
Jump to
Diff view
Diff view
There are no files selected for viewing
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,24 @@ | ||
| # TC: O(amount * len(coins)) | ||
| # SC: O(amount) | ||
| class Solution: | ||
| def coinChange(self, coins: List[int], amount: int) -> int: | ||
|
|
||
| from collections import deque | ||
|
|
||
| v = {0: 0} | ||
| q = deque([0]) | ||
|
|
||
| while q and amount not in v: | ||
|
|
||
| cur = q.popleft() | ||
|
|
||
| for coin in coins: | ||
|
|
||
| if cur + coin > amount or cur + coin in v: | ||
| continue | ||
|
|
||
| v[cur + coin] = v[cur] + 1 | ||
| q.append(cur + coin) | ||
|
|
||
| return -1 if amount not in v else v[amount] | ||
|
|
|
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. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 양 끝과 중간 값을 비교해 최소 원소의 위치를 이진 탐색으로 좁혀 나갑니다. 개선 제안: 현재 구현이 적절해 보입니다.
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. 저는 while loop 안에 복잡한 연산 들어가는거 별로라 최대한 간단하게 했는데, 이렇게 하시면 얼리 리턴이 가능해서 좋네요! |
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -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 | ||
|
|
||
| while low < high - 1: | ||
| mid = (low + high) // 2 | ||
|
|
||
| if nums[low] < nums[mid] < nums[high]: | ||
| return nums[low] | ||
|
|
||
| if nums[low] > nums[mid]: | ||
| high = mid | ||
| else: | ||
| low = mid | ||
|
|
||
| return min(nums[low], nums[high]) | ||
|
|
|
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. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 트리의 모든 노드를 방문하며 재귀 스택으로 깊이를 추적합니다. 개선 제안: 현재 구현이 적절해 보입니다. |
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -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: | ||
|
|
||
| def rec(node, height): | ||
| if not node: | ||
| return height | ||
|
|
||
| return max(rec(node.left, height + 1), rec(node.right, height + 1)) | ||
|
|
||
| return rec(root, 0) | ||
|
|
|
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!
Contributor
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. 의견 감사합니다.
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. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 둘 다 정렬되어 있으므로 더 작은 노드를 순차적으로 연결합니다. 개선 제안: 현재 구현이 적절해 보입니다. |
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -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]: | ||
|
|
||
| ret = ListNode() | ||
| cur = ret | ||
|
|
||
| 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 | ||
|
|
||
| cur.next = list1 if list1 else list2 | ||
|
|
||
| return ret.next | ||
|
|
|
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. 🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 각 위치에서 상하좌좌로 탐색하며 단어 길이만큼 가지치를 수행합니다. 개선 제안: 현재 구현이 적절해 보입니다.
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! |
| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,30 @@ | ||
| # TC: O(4^L) | ||
| # SC: O(L) | ||
| class Solution: | ||
| def exist(self, board: List[List[str]], word: str) -> bool: | ||
|
|
||
| n = len(board) | ||
| m = len(board[0]) | ||
|
|
||
| def rec(row, col, idx): | ||
| if idx == len(word): | ||
| return True | ||
|
|
||
| if row < 0 or row >= n or col < 0 or col >= m or board[row][col] != word[idx]: | ||
| return False | ||
|
|
||
| board[row][col] = None | ||
|
|
||
| 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) | ||
|
|
||
| board[row][col] = word[idx] | ||
|
|
||
| return ret | ||
|
|
||
| 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 | ||
|
|
||
| return False | ||
|
|
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
🏷️ 알고리즘 패턴 분석
📊 시간/공간 복잡도 분석
피드백: 도달 가능한 합을 층별로 확장하는 BFS 방식으로 최단 거리를 보장하여 총합 만큼의 상태 공간을 방문합니다.
개선 제안: 현재 구현이 적절해 보입니다.
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
저는 코인이 양수값만 있어서 DP만 했었는데 BFS로 해도 충분히 가능하겠네요!