LeetCode 79. Word Search (javascript)

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.

Idea:

DFS

Solution:

/**
 * @param {character[][]} board
 * @param {string} word
 * @return {boolean}
 */
var exist = function(board, word) {
    for (let x = 0; x < board[0].length; x++)
        for (let y = 0; y < board.length; y++)
            if (dfs(board, word, 0, x, y)) return true;
    return false;
};

function dfs(board, word, i, x, y) {
    // exit recursion:
    if (x < 0 || y < 0 || x >= board[0].length || y >= board.length || word[i] !== board[y][x])
        return false;
    // found the last letter
    if (i === word.length - 1)
        return true;
    
    let cur = board[y][x];
    // mark it 0 means already visit
    board[y][x] = 0;
    let found = dfs(board, word, i + 1, x + 1, y) || 
                dfs(board, word, i + 1, x - 1, y) ||
                dfs(board, word, i + 1, x, y + 1) ||
                dfs(board, word, i + 1, x, y - 1);
    // recover
    board[y][x] = cur;
    return found;
}