Minesweeper Python Solution with Tests

Tested Python solution for LeetCode 529 with 20 pytest cases. Generate a practice environment with lcpy.

LeetCode 529, Medium. Topics: Array, Depth-First Search, Breadth-First Search, Matrix. View on LeetCode.

Generate this problem as a practice environment: tested reference solution, 20 parametrized pytest cases, and a playground notebook:

lcpy gen -n 529   # by problem number
lcpy gen -s minesweeper   # by problem name

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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.

You are also given an integer array click where click = [clickr, clickc] represents the next click position among all the unrevealed squares ('M' or 'E').

Return the board after revealing this position according to the following rules:

1. If a mine 'M' is revealed, then the game is over. You should change it to 'X'.
2. If an empty square 'E' with no adjacent mines is revealed, then change it to a revealed blank 'B' and all of its adjacent unrevealed squares should be revealed recursively.
3. If an empty square 'E' with at least one adjacent mine is revealed, then change it to a digit ('1' to '8') representing the number of adjacent mines.
4. Return the board when no more squares will be revealed.

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Examples

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"]]

Input: board = [["B","1","E","1","B"],["B","1","M","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]], click = [1,2]
Output: [["B","1","E","1","B"],["B","1","X","1","B"],["B","1","1","1","B"],["B","B","B","B","B"]]

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Constraints

• m == board.length
• n == board[i].length
• 1 <= m, n <= 50
• board[i][j] is either 'M', 'E', 'B', or a digit from '1' to '8'.
• click.length == 2
• 0 <= clickr < m
• 0 <= clickc < n
• board[clickr][clickc] is either 'M' or 'E'.

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Solution

Reference implementation from solution.py on GitHub, full suite in test_solution.py:

from collections import deque


class Solution:
    DIRECTIONS: tuple[tuple[int, int], ...] = (
        (-1, -1),
        (-1, 0),
        (-1, 1),
        (0, -1),
        (0, 1),
        (1, -1),
        (1, 0),
        (1, 1),
    )

    # Time: O(m * n)
    # Space: O(m * n)
    def update_board(self, board: list[list[str]], click: list[int]) -> list[list[str]]:
        rows, cols = len(board), len(board[0])
        row, col = click
        if board[row][col] == "M":
            board[row][col] = "X"
            return board

        queue: deque[tuple[int, int]] = deque([(row, col)])
        while queue:
            r, c = queue.popleft()
            mines = self._adjacent_mines(board, r, c)
            if mines:
                board[r][c] = str(mines)
                continue
            board[r][c] = "B"
            for dr, dc in self.DIRECTIONS:
                nr, nc = r + dr, c + dc
                if 0 <= nr < rows and 0 <= nc < cols and board[nr][nc] == "E":
                    board[nr][nc] = "B"
                    queue.append((nr, nc))
        return board

    def _adjacent_mines(self, board: list[list[str]], r: int, c: int) -> int:
        rows, cols = len(board), len(board[0])
        return sum(
            1
            for dr, dc in self.DIRECTIONS
            if 0 <= r + dr < rows and 0 <= c + dc < cols and board[r + dr][c + dc] == "M"
        )

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Complexity

Time	Space
O(m * n)	O(m * n)

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