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[11/24] 79. Word Search

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작성일 2022.11.24 09:44

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Medium

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Given an m x n grid of characters board and a string word, return true if word exists in the grid.

The word can be constructed from letters of sequentially adjacent cells, where adjacent cells are horizontally or vertically neighboring. The same letter cell may not be used more than once.

Example 1:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
Output: true

Example 2:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
Output: true

Example 3:

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
Output: false

Constraints:

m == board.length
n = board[i].length
1 <= m, n <= 6
1 <= word.length <= 15
board and word consists of only lowercase and uppercase English letters.

Follow up: Could you use search pruning to make your solution faster with a larger board?

Accepted

1,173,744

Submissions

2,929,480

#Amazon, #Bloomberg, #Uber, #Karat, #Microsoft, #Cisco, #Google, #TikTok, #Indeed, #Facebook, #Apple, #Walmart Global Tech, #Adobe

링크

https://leetcode.com/problems/word-search/ 2296 회 연결

익명
작성일 2022.11.24 09:44

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        ROW, COL = len(board), len(board[0])
        visited = set()
        def dfs(row, col, i):
            
            if (row >= ROW or col >= COL or row < 0 or col < 0 or (row,col) in visited or word[i] != board[row][col]):
                return False
            visited.add((row,col))
            if i == len(word)-1:
                return True
            
            ret_val = dfs(row+1, col, i+1) or dfs(row, col+1, i+1) or dfs(row-1,col,i+1) or dfs(row, col-1, i+1)
            
            visited.remove((row,col))
            return ret_val
        
        for row in range(ROW):
            for col in range(COL):
                if dfs(row, col, 0): return True
        return False