0079. Word Search

https://leetcode.com/problems/word-search

Description

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?

ac

class Solution {
    int[][] dirs = new int[][]{{0,1},{0,-1},{1,0},{-1,0}};

    public boolean exist(char[][] board, String word) {
        // edge cases
        if (board == null || board.length == 0 || board[0].length == 0 || word.length() == 0) {
            return false;
        }

        for (int r = 0; r < board.length; r++) {
            for (int c = 0; c < board[0].length; c++) {
                if (backtrack(board, r, c, 0, word)) return true;
            }
        }

        return false;
    }

    private boolean backtrack(char[][] board, int r, int c, int kth, String word) {
        // exit
        if (r < 0 || r >= board.length || c < 0 || c >= board[0].length || board[r][c] != word.charAt(kth))
            return false;
        if (kth == word.length()-1) return true;

        char old = board[r][c];
        board[r][c] = '*';

        for (int[] d : dirs) {
            int r2 = r + d[0];
            int c2 = c + d[1];
            if (backtrack(board, r2, c2, kth+1, word)) return true; // find one true, return, no need to proceed
        }

        board[r][c] = old;

        return false;
    }
}