Last updated on October 5th, 2024 at 06:04 pm

Here, We see Minesweeper LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Minesweeper LeetCode Solution

Problem Statement

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”]]

1. Minesweeper LeetCode Solution C++

class Solution {
public:
    int dx[8] = {-1, 0, 1, 0, -1, -1, 1, 1};
    int dy[8] = {0, 1, 0, -1, -1, 1, 1, -1};
    bool isValid(vector<vector<char>>& board, int i, int j) {
        return (i >= 0 && j >= 0 && i < board.size() && j < board[0].size());
    }
    
    int hasAdjacentMine(vector<vector<char>>& board, int i, int j) {
        int count = 0;
        for (int k=0; k<8; k++) {
            int I = i + dx[k];
            int J = j + dy[k];
            if (isValid(board, I, J) && board[I][J] == 'M')
                count++;
        }
        return count;
    }
    
    void dfs(vector<vector<char>>& board, vector<vector<bool>>& visited, int i, int j) {
        if (min(i, j) < 0 || i >= board.size() || j >= board[0].size() || visited[i][j]) return;
        visited[i][j] = true;
        if (board[i][j] == 'M') {
            board[i][j] = 'X';
            return;
        }
        if (board[i][j] == 'E') {
            int c = hasAdjacentMine(board, i, j);
            if (c == 0) {
                board[i][j] = 'B';
                for (int k=0; k<8; k++) {
                    dfs(board, visited, i+dx[k], j+dy[k]);
                }
            }
            else {
                board[i][j] = c + '0';
                return;
            }
        }
    }
    
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        vector<vector<bool>> visited(board.size(), vector<bool>(board[0].size(), false));
        dfs(board, visited, click[0], click[1]);
        return board;
    }
};

2. Minesweeper LeetCode Solution Java

class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        if (board[click[0]][click[1]] == 'M') {
            board[click[0]][click[1]] = 'X';
            return board;
        }
        reveal(board, click[0], click[1]);
        return board;
    }
    
    private void reveal(char[][] board, int i, int j) {
        if (i < 0 || j < 0 || i >= board.length || j >= board[0].length || board[i][j] != 'E')
            return;
        board[i][j] = '0';
        int[][] neighbors = {{i-1, j-1}, {i-1, j}, {i-1, j+1}, 
                             {i, j-1}, {i, j+1}, 
                             {i+1, j-1}, {i+1, j}, {i+1, j+1}};
        for (int[] neighbor : neighbors) {
            if (neighbor[0] < 0 || neighbor[1] < 0 || neighbor[0] >= board.length || neighbor[1] >= board[0].length)
                continue;
            if (board[neighbor[0]][neighbor[1]] == 'M')
                board[i][j] ++;
        }
        if (board[i][j] != '0')
            return;
        board[i][j] = 'B';
        for (int[] neighbor : neighbors)
            reveal(board, neighbor[0], neighbor[1]);
    }
}

3. Minesweeper LeetCode Solution JavaScript

var updateBoard = function(board, click) {
  const rows = board.length;
  const cols = board[0].length;
  dfs(click[0], click[1]);
  return board;
  function dfs(i, j) {
    if (!board[i][j]) return;
    if (board[i][j] === 'M') {
      board[i][j] = 'X';
      return;
    }
    if (board[i][j] !== 'E') return;
    const mines = checkForMine(i, j);
    if (mines) {
      board[i][j] = mines.toString();
      return;
    } else {
      board[i][j] = 'B';
      for (let x = Math.max(i - 1, 0); x < Math.min(i + 2, rows); x++) {
        for (let y = Math.max(j - 1, 0); y < Math.min(j + 2, cols); y++) {
          dfs(x, y);
        }
      }
    }
  }

  function checkForMine(x, y) {
    let mines = 0;
    for (let i = Math.max(x - 1, 0); i < Math.min(x + 2, rows); i++) {
      for (let j = Math.max(y - 1, 0); j < Math.min(y + 2, cols); j++) {
        if (board[i][j] === 'M') mines++;
      }
    }
    return mines;
  }
}

4. Minesweeper LeetCode Solution Python

class Solution(object):
    def updateBoard(self, board, click):
        def adjacent_mines(i, j):
            mines = 0
            for x, y in directions:
                if 0 <= i + x < M and 0 <= j + y < N and board[i + x][j + y] == "M":
                    mines += 1
            return mines
        def dfs(i, j):
            mines = adjacent_mines(i, j)
            if mines:
                board[i][j] = str(mines)
                return board
            board[i][j] = "B"
            for x, y in directions:
                if 0 <= i + x < M and 0 <= j + y < N and board[i + x][j + y] == "E":
                    dfs(i + x, j + y)
            return board
        if not board: return board
        M = len(board)
        N = len(board[0])
        directions = [(0, 1), (1, 0), (0, -1), (-1, 0), (1,1), (1,-1), (-1,1), (-1,-1)]
        if board[click[0]][click[1]] == "M":
            board[click[0]][click[1]] = "X"
            return board
        if board[click[0]][click[1]] == "E":
            return dfs(click[0], click[1])