LeetCode Minesweeper Solution

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.

Example 1:

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"]]

Example 2:

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"]]

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'.

Problem Link: https://leetcode.com/problems/minesweeper/



Solution

There are a few things we need to keep in mind when implementing the problem:

  • blank squares ('B') and digit squares ('1'-'8') should be treated differently
  • a blank square's neighbours need to be revealed recursively
  • only recursively reveal empty squares ('E')

We are given the click coordinate to update the board. If click is a mine ('M'), then we change it to 'X'. If it's an empty square, then the update function updates the empty square recursively.

Implementation

1def updateBoard(board: List[List[str]], click: List[int]) -> List[List[str]]:
2    if board == []: return []
3    row = len(board)
4    col = len(board[0])
5
6    def isbomb(i, j):
7        if i < 0 or j < 0 or i >= row or j >= col:
8            return False
9        elif board[i][j] == 'M' or board[i][j] == 'X':
10            return True
11        else: return False
12    
13    def update(r, c):
14        if r < 0 or c < 0 or r >= row or c >= col: return
15        if board[r][c] == 'E':    # empty square
16            bombs = 0             # number of neighbouring bombs
17            for i in range(r-1, r+2):
18                for j in range(c-1, c+2):
19                    if isbomb(i, j): bombs += 1
20            if bombs:             # update cell with number of bombs
21                board[r][c] = chr(48 + bombs)
22            else:                 # blank cell (0 bombs)
23                board[r][c] = 'B'
24                for i in range(r-1, r+2): # update neighbours
25                    for j in range(c-1, c+2):
26                        update(i, j)
27        # else already revealed, do nothing
28    
29    r, c = click[0], click[1]
30    if board[r][c] == 'M':  # bomb
31        board[r][c] = 'X'
32    else:                   # empty square, or revealed square
33        update(r, c)
34    return board