79 Word Search

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:

(A) (B) (C)  E
 S   F  (C)  S
 A  (D) (E)  E

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],
word = "ABCCED"
Output: true

Example 2:

 A   B   C   E
 S   F   C  (S)
 A   D  (E) (E)

Input: board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]],
word = "SEE"
Output: true

Example 3:

 A   B   C   E
 S   F   C   S
 A   D   E   E

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?

Try backtracking (DFS) recursively from each cell.

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class Solution:
    def exist(self, board: List[List[str]], word: str) -> bool:
        self.board = board
        for i in range(len(board)):
            for j in range(len(board[0])):
                if self.backtrack(i, j, word):
                    return True
        return False
        
    def backtrack(self, i: int, j: int, suffix: str) -> bool:
        if len(suffix) == 0:
            return True
        
        if (
            i < 0 or i >= len(self.board)
            or j < 0 or j >= len(self.board[0])
            or self.board[i][j] != suffix[0]
        ):
            return False
        
        result = False
        self.board[i][j] = '#'
        for offset in [(-1, 0), (1, 0), (0, -1), (0, 1)]:
            result = self.backtrack(i+offset[0], j+offset[1], suffix[1:])
            if result:
                break # not return here for cleanup side-effects
        self.board[i][j] = suffix[0]
        return result