Minesweeper

BFS

Problem

Let's play the minesweeper game (Wikipedia, online game)!

You are given a 2D char matrix representing the game board. 'M' represents an unrevealed mine, 'E' represents an unrevealed empty square, 'B' represents a revealed blank square that has no adjacent (above, below, left, right, and all 4 diagonals) mines, digit ('1' to '8') represents how many mines are adjacent to this revealed square, and finally 'X' represents a revealed mine.

Now given the next click position (row and column indices) 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 - change it to 'X'.

2. If an empty square ('E') with no adjacent mines is revealed, then change it to 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.

Thought Process

• There's a BFS solution and DFS solution. I implemented both

Solution

from collections import deque

class Solution:
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        
        visited = set()
        q = deque()
        i, j = click[0], click[1]
        q.append([i,j])
        visited.add((i,j))
        
        directions = [[-1,0], [1,0], [0,-1], [0,1],[-1,-1], [-1,1], [1,-1], [1,1]]
        
        while q:
            coor = q.popleft()
            print(q)
            x, y = coor[0], coor[1]
            print(x,y)
        
            if board[x][y] == 'M':
                board[x][y] = 'X'
            
            else: #see how many bombs are around
                
                count = 0
                for d in directions:
                    newX, newY = x+d[0], y+d[1]
                  
                    
                    if newX < 0 or newX >= len(board) or newY < 0 or newY >= len(board[0]):
                        continue
                    elif board[newX][newY] == 'M' or board[newX][newY] == 'X' :
                        count+=1
                        
                if count != 0: #we don't want to do BFS if a bomb is near
                    board[x][y] = str(count)
                    
                else:
                    board[x][y] = 'B'
                    
                    for dd in directions:
                        newX, newY = x+dd[0], y+dd[1]
                        if newX < 0 or newX >= len(board) or newY < 0 or newY >= len(board[0]) or (newX, newY) in visited:
                            continue
                        
                        if board[newX][newY] == 'E':
                            q.append([newX, newY])
                            
                            visited.add((newX, newY))
            
        return board
                        

DFS Solution

class Solution:
    def checkMines(self, x, y, board):
        numMines = 0
        for i in range(x-1,x+2):
            for j in range(y-1, y+2):
                if i >= 0 and i < len(board) and j >= 0 and j < len(board[i]) and board[i][j] == 'M':
                    numMines+=1
        return numMines
        
    def updateBoard(self, board: List[List[str]], click: List[int]) -> List[List[str]]:
        
        x, y = click[0], click[1]
        
        if board[x][y] == 'M':
            board[x][y] = 'X'
              
        else:
            mines = self.checkMines(x,y,board)
            if mines:
                board[x][y] = str(mines)
            else: #no mines adjacent
                board[x][y] = 'B'
                for i in range(x-1,x+2):
                    for j in range(y-1, y+2):
                        if i >= 0 and i < len(board) and j >= 0 and j < len(board[i]) and board[i][j] != 'B':
                            self.updateBoard(board, [i,j])
        return board
        

Time Complexity

• Time: O(m*n) for both BFS and DFS since we are traversing through the matrix

• Space: O(m*n) for both BFS and DFS. With a queue we are storing every cell and with recusrion the call stack will be m*n.