Word Search

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.

For example,
Given board =

[
[‘A’,’B’,’C’,’E’],
[‘S’,’F’,’C’,’S’],
[‘A’,’D’,’E’,’E’]
]
word = “ABCCED”, -> returns true,
word = “SEE”, -> returns true,
word = “ABCB”, -> returns false.

public class Solution {
    public boolean exist(char[][] board, String word) {
        int m = board.length; 
        int n = board[0].length; 
        boolean result = false; 
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (dfs(board, word, i, j, 0)) {
                    result = true; 
                }
            }
        }
        return result; 
        
    }
    
    public boolean dfs (char[][] board, String word, int i, int j, int k) {
        int m = board.length; 
        if (m == 0) return false; 
        int n = board[0].length; 
        if ( i < 0 || j < 0 || i >= m || j >= n) return false; 
        if (board[i][j] == word.charAt(k)) {
            char temp = board[i][j]; 
            board[i][j] = '#'; //have been used
            if(k == word.length()-1) return true;
            else if (dfs(board, word, i-1, j, k+1) 
                || dfs(board, word, i+1, j, k+1)
                || dfs(board, word, i, j-1, k+1)
                || dfs(board, word, i, j+1, k+1)) {
                return true; 
            }
            board[i][j] = temp; 
        }
        return false; 
    }
}
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