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79. Word Search

Difficulty Topics

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?

Solution

word-search.py
class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        rows, cols = len(board), len(board[0])
        N = len(word)
        visited = [[False] * cols for _ in range(rows)]

        def go(x, y, index):
            if index == N: return True

            for dx, dy in [(x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)]:
                if 0 <= dx < rows and 0 <= dy < cols and not visited[dx][dy] and board[dx][dy] == word[index]:
                    visited[dx][dy] = True
                    if go(dx, dy, index + 1): return True
                    visited[dx][dy] = False

            return False

        for i in range(rows):
            for j in range(cols):
                if board[i][j] == word[0]:
                    visited[i][j] = True
                    if go(i, j, 1): return True
                    visited[i][j] = False

        return False