LiDARism

111. Minimum Depth of Binary Tree

https://leetcode.com/problems/minimum-depth-of-binary-tree/description/ 최단 거리 구하기는 BFS로 하는 것이 정석이다.# 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 = rightclass Solution: def minDepth(self, root: Optional[TreeNode]) -> int: if root is None: retur..

101. Symmetric Tree

https://leetcode.com/problems/symmetric-tree/ BFS를 대칭에 맞도록... left right를 따로따로 해주자 # 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 = rightclass Solution: def isSymmetric(self, root: Optional[TreeNode]) -> bool: if not root: return True q..

1138. Alphabet Board Path

https://leetcode.com/problems/alphabet-board-path/description/ 그냥... 그냥 풀면 된다.하지만 난 못풀었지 class Solution: def alphabetBoardPath(self, target: str) -> str: r = c = 0 out = [] for ch in target: tr, tc = divmod(ord(ch) - ord('a'), 5) dr, dc = tr - r, tc - c if dr 0: out.append('D' * dr) if dc > 0: o..

https://leetcode.com/problems/shortest-bridge/description/ BFS/DFS의 queue/stack 내부에 counter를 넣어서 문제를 해결할 수 있다. class Solution: def shortestBridge(self, grid: List[List[int]]) -> int: m = len(grid) n = len(grid[0]) dirs = [(1, 0), (-1, 0), (0, 1), (0, -1)] ## directions? vis = [[False]*n for _ in range(m)] ## visited ??? queue = [] def dfs(r, c): ..

https://leetcode.com/problems/detonate-the-maximum-bombs/ adjacent를 표현하는 그래프를 먼저 그리고 그에 맞도록 DFS를 해주면 된다. 그리고 한 칸 이동 = stack pop 할 때마다 카운터 +1 하면 된다. class Solution: def maximumDetonation(self, bombs: List[List[int]]) -> int: n = len(bombs) adj = [[] for _ in range(n)] # Make Graph: i can bomb out j? for i in range(n): xi, yi, ri = bombs[i] r..

https://leetcode.com/problems/surrounded-regions/ 테두리에 있는 거 찾아서 제거하는 게 차라리 낫다. BFSclass Solution: def solve(self, board: List[List[str]]) -> None: """ Do not return anything, modify board in-place instead. """ if not board or not board[0]: return m, n = len(board), len(board[0]) def bfs(r: int, c: int) -> None: queue = [(r, c)] ..

https://leetcode.com/problems/flood-fill/?utm_source=chatgpt.com BFS로 단순하게 풀 수 있다.조건만 잘 설정하면... class Solution: def floodFill(self, image: List[List[int]], sr: int, sc: int, color: int) -> List[List[int]]: dirs = [(1,0),(-1,0),(0,1),(0,-1)] m, n = len(image), len(image[0]) starting_color = image[sr][sc] vis = [[0]*n for _ in range(m)] queue = [(sr, sc)..

https://leetcode.com/problems/map-of-highest-peak/description/ 최단 거리랑 유사한 문제라 BFS를 하면 된다. class Solution: def highestPeak(self, isWater: List[List[int]]) -> List[List[int]]: m, n = len(isWater), len(isWater[0]) height = [[-1]*n for _ in range(m)] queue = [] ## set water to 0. start in water for i in range(m): for j in range(n): if..

https://leetcode.com/problems/count-servers-that-communicate/description/ DFS / BFS 아니어도 풀 수는 있다. class Solution: def countServers(self, grid: List[List[int]]) -> int: m, n = len(grid), len(grid[0]) row_cnt = [0]*m col_cnt = [0]*n # 1) 행/열별 서버 수 카운트 for i in range(m): for j in range(n): if grid[i][j] == 1: ..

https://leetcode.com/problems/merge-k-sorted-lists/ You are given an array of k linked-lists lists, each linked-list is sorted in ascending order.Merge all the linked-lists into one sorted linked-list and return it. Example 1:Input: lists = [[1,4,5],[1,3,4],[2,6]]Output: [1,1,2,3,4,4,5,6]Explanation: The linked-lists are:[ 1->4->5, 1->3->4, 2->6]merging them into one sorted linked list:1->1->..