Minesweeper — LeetCode 529 Python Solution

The problem

Let's play the minesweeper game (Wikipedia, online game)! You are given an m x n char matrix board representing the game board where: 'M' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent mines (i.e., above, below, left, right, and all 4 diagonals), digit ('1' to '8') represents how many mines are adjacent to this revealed square, and 'X' represents a revealed mine.

Example

Input

board = [["E","E","E","E","E"],["E","E","M","E","E"],["E","E","E","E","E"],["E","E","E","E","E"]], click = [3,0]

Output

[["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]

Python solution

class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        def dfs(i: int, j: int):
            cnt = 0
            for x in range(i - 1, i + 2):
                for y in range(j - 1, j + 2):
                    if 0 <= x < m and 0 <= y < n and board[x][y] == "M":
                        cnt += 1
            if cnt:
                board[i][j] = str(cnt)
            else:
                board[i][j] = "B"
                for x in range(i - 1, i + 2):
                    for y in range(j - 1, j + 2):
                        if 0 <= x < m and 0 <= y < n and board[x][y] == "E":
                            dfs(x, y)

        m, n = len(board), len(board[0])
        i, j = click
        if board[i][j] == "M":
            board[i][j] = "X"
        else:
            dfs(i, j)
        return board

Complexity

Pattern: Matrix and Grid

Treat a 2-D grid as a graph whose neighbours are the four adjacent cells. LeetCode 529. Minesweeper is filed here because LeetCode tags it Matrix, which is the vocabulary this hub collects.

The matrix and grid guide has the Python template for the pattern and the 216 LeetCode problems that use it.

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