79. Word Search

79. Word Search
Medium

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Given a 2D board and a word, find if the word exists in the grid.

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

Example:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

Given word = "ABCCED", return true.
Given word = "SEE", return true.
Given word = "ABCB", return false.



class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.dfs(board, i, j, word):
                    return True
        return False



    def dfs(self, board, i, j, word):
        if len(word) == 0:
            return True


        if i<0 or j<0 or i>len(board)-1 or j>len(board[0])-1 or board[i][j]!=word[0]:
            return
        temp = board[i][j]
        board[i][j] = '#'
        res = self.dfs(board, i-1, j, word[1:]) or self.dfs(board, i+1, j, word[1:]) or \
                self.dfs(board, i, j-1, word[1:]) or self.dfs(board, i, j+1, word[1:])
        board[i][j] = temp

        return res