Word search

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:


        found = False
        rows, cols = len(board), len(board[0])
        used = set()
        def dfs(i, j, index):
            nonlocal found
            if index == len(word):
                found = True
                return
            if found:
                return
            if i < 0 or j < 0 or i >= rows or j >= cols:
                return
            if board[i][j] != word[index]:
                return
            if (i, j) in used:
                return

            used.add((i, j))

            for drow, dcol in [(1, 0), (0, 1), (-1, 0), (0, -1)]:
                dfs(i - drow, j - dcol, index + 1)

            used.remove((i, j))

        for i in range(rows):
            for j in range(cols):
                dfs(i, j, 0)
                if found:
                    return True
        return False

Word Search

Difficulty: Medium

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?