[Leetcode] 0079. Word Search

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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?

內容目錄

Python

				
					# time complexity: O(c*3^l)
# space complexity: O(l)
from typing import List


class Solution:

    def exist(self, board: List[List[str]], word: str) -> bool:
        def backtrack(suffix: str, r: int, c: int):
            if len(suffix) == 0:
                return True
            if not (0 <= r < ROW and 0 <= c < COL) or suffix[0] != board[r][c]:
                return False
            result = False
            originalChar = board[r][c]
            board[r][c] = "#"
            for dr,dc in ([1,0],[0,1],[-1,0],[0,-1]):
                result = backtrack(suffix[1:], r+dr, c+dc)
                if result:
                    break
            board[r][c] = originalChar
            return result
            
        ROW = len(board)
        COL = len(board[0])
        for row in range(ROW):
            for col in range(COL):
                if backtrack(word, row, col):
                    return True
        return False
        


board = [["A", "B", "C", "E"], ["S", "F", "C", "S"], ["A", "D", "E", "E"]]
word = "ABCCED"

print(Solution().exist(board, word))
				
			
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