README_EN.md

79. Word Search

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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?

Solutions

Python3

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        def dfs(i, j, cur):
            if cur == len(word):
                return True
            if i < 0 or i >= m or j < 0 or j >= n or board[i][j] == '0' or word[cur] != board[i][j]:
                return False
            t = board[i][j]
            board[i][j] = '0'
            for a, b in [[0, 1], [0, -1], [-1, 0], [1, 0]]:
                x, y = i + a, j + b
                if dfs(x, y, cur + 1):
                    return True
            board[i][j] = t
            return False

        m, n = len(board), len(board[0])
        return any(dfs(i, j, 0) for i in range(m) for j in range(n))

Java

class Solution {
    public boolean exist(char[][] board, String word) {
        int m = board.length;
        int n = board[0].length;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (dfs(i, j, 0, m, n, board, word)) {
                    return true;
                }
            }
        }
        return false;
    }

    private boolean dfs(int i, int j, int cur, int m, int n, char[][] board, String word) {
        if (cur == word.length()) {
            return true;
        }
        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word.charAt(cur)) {
            return false;
        }
        board[i][j] += 256;
        int[] dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k) {
            int x = i + dirs[k];
            int y = j + dirs[k + 1];
            if (dfs(x, y, cur + 1, m, n, board, word)) {
                return true;
            }
        }
        board[i][j] -= 256;
        return false;
    }
}

TypeScript

function exist(board: string[][], word: string): boolean {
    let m = board.length,
        n = board[0].length;
    let visited = Array.from({ length: m }, v => new Array(n).fill(false));
    for (let i = 0; i < m; ++i) {
        for (let j = 0; j < n; ++j) {
            if (dfs(board, word, i, j, 0, visited)) {
                return true;
            }
        }
    }
    return false;
}

function dfs(
    board: string[][],
    word: string,
    i: number,
    j: number,
    depth: number,
    visited: boolean[][],
): boolean {
    let m = board.length,
        n = board[0].length;
    if (i < 0 || i > m - 1 || j < 0 || j > n - 1 || visited[i][j]) {
        return false;
    }
    if (board[i][j] != word.charAt(depth)) {
        return false;
    }

    if (depth == word.length - 1) {
        return true;
    }

    visited[i][j] = true;
    ++depth;
    let res = false;
    for (let [dx, dy] of [
        [0, 1],
        [0, -1],
        [1, 0],
        [-1, 0],
    ]) {
        let x = i + dx,
            y = j + dy;
        res = res || dfs(board, word, x, y, depth, visited);
    }
    visited[i][j] = false;
    return res;
}

C++

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        int m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i)
            for (int j = 0; j < n; ++j)
                if (dfs(i, j, 0, m, n, board, word))
                    return true;
        return false;
    }

    bool dfs(int i, int j, int cur, int m, int n, vector<vector<char>>& board, string& word) {
        if (cur == word.size()) return true;
        if (i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[cur]) return false;
        char t = board[i][j];
        board[i][j] = '0';
        vector<int> dirs = {-1, 0, 1, 0, -1};
        for (int k = 0; k < 4; ++k)
        {
            int x = i + dirs[k], y = j + dirs[k + 1];
            if (dfs(x, y, cur + 1, m, n, board, word)) return true;
        }
        board[i][j] = t;
        return false;
    }
};

Go

func exist(board [][]byte, word string) bool {
	m, n := len(board), len(board[0])
	var dfs func(i, j, cur int) bool
	dfs = func(i, j, cur int) bool {
		if cur == len(word) {
			return true
		}
		if i < 0 || i >= m || j < 0 || j >= n || board[i][j] != word[cur] {
			return false
		}
		t := board[i][j]
		board[i][j] = '0'
		dirs := []int{-1, 0, 1, 0, -1}
		for k := 0; k < 4; k++ {
			x, y := i+dirs[k], j+dirs[k+1]
			if dfs(x, y, cur+1) {
				return true
			}
		}
		board[i][j] = t
		return false
	}
	for i := 0; i < m; i++ {
		for j := 0; j < n; j++ {
			if dfs(i, j, 0) {
				return true
			}
		}
	}
	return false
}

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