# [LC] 79. 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.

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.Time: O(M * N * 4^|Word|)```
``` 1 class Solution {
2     int row;
3     int col;
4     public boolean exist(char[][] board, String word) {
5         row = board.length;
6         col = board[0].length;
7         boolean[][] visited = new boolean[row][col];
8         for (int i = 0; i < row; i++) {
9             for (int j = 0; j < col; j++) {
10                 if (helper(board, visited, word, i, j, 0)) {
11                     return true;
12                 }
13             }
14         }
15         return false;
16     }
17
18     private boolean helper(char[][] board, boolean[][] visited, String word, int i, int j, int index) {
19         if (index == word.length()) {
20             return true;
21         }
22         if (i < 0 || i >= row || j < 0 || j >= col) {
23             return false;
24         }
25         if (word.charAt(index) == board[i][j] && !visited[i][j]) {
26             visited[i][j] = true;
27             boolean res = helper(board, visited, word, i + 1, j, index + 1) ||
28                 helper(board, visited, word, i - 1, j, index + 1) ||
29                 helper(board, visited, word, i, j + 1, index + 1) ||
30                 helper(board, visited, word, i, j - 1, index + 1);
31             // need to clean up visited
32             visited[i][j] = false;
33             return res;
34         }
35
36         return false;
37     }
38 }```

[LC] 79. Word Search

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