Word Search

Problem

https://leetcode.com/problems/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:

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?

Pattern

Array, String, Backtracking, Depth-First Search, Matrix

Approaches

Backtracking

Code

def exist(board: list[list[str]], word: str) -> bool:
    M = len(board)
    N = len(board[0])

    def backtrack(w, path, i, j):

        if (i, j) in path or not (0 <= i < M and 0 <= j < N):
            return False

        path.add((i, j))

        if board[i][j] != word[w - 1]:
            return False

        if w == len(word):
            return True

        return any(
            [
                backtrack(w + 1, path.copy(), i - 1, j),
                backtrack(w + 1, path.copy(), i + 1, j),
                backtrack(w + 1, path.copy(), i, j - 1),
                backtrack(w + 1, path.copy(), i, j + 1),
            ]
        )

    for i in range(M):
        for j in range(N):
            if board[i][j] == word[0] and backtrack(1, set(), i, j):
                return True

    return False

Test

>>> from word_search__backtracking import exist
>>> exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCCED")
True
>>> exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "SEE")
True
>>> exist([["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], "ABCB")
False

word_search__backtracking.exist(board: list[list[str]], word: str) → bool