word_search.py

"""
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?
"""

class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        m = len(board)
        n = len(board[0])

        def dfs(i, j, cur): # cur: current char count
            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' # mark as visited

            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
        
        # Initiate DFS from every cell in the board
        for i in range(m):
            for j in range(n):
                if board[i][j] == word[0] and dfs(i, j, 0):
                    return True
        
        return False