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LeetCode 529: Minesweeper Solution

Master LeetCode problem 529 (Minesweeper), a medium challenge, with our optimized solutions in Java, C++, and Python. Explore detailed explanations, test your code in our interactive editor, and prepare for coding interviews.

Problem Explanation

Explanation:

To solve this problem, we can use a depth-first search (DFS) approach to reveal squares on the board based on the rules provided. We will start by checking the click position and its neighboring squares to determine the appropriate action for each square. We will update the board accordingly until no more squares can be revealed.

Algorithm:

  1. If the click position is a mine, change it to 'X' and return the board.
  2. If the click position is an empty square with no adjacent mines, change it to 'B' and recursively reveal all adjacent unrevealed squares by calling the DFS function.
  3. If the click position is an empty square with adjacent mines, change it to the number of adjacent mines.
  4. In the DFS function, for each unrevealed square adjacent to the current square, repeat steps 2 and 3.

Time Complexity:

The time complexity of this algorithm is O(m*n) where m is the number of rows and n is the number of columns in the board.

Space Complexity:

The space complexity of this algorithm is O(m*n) due to the recursive calls in the DFS function.

:

Solution Code

class Solution {
    public char[][] updateBoard(char[][] board, int[] click) {
        int row = click[0];
        int col = click[1];
        
        if (board[row][col] == 'M') {
            board[row][col] = 'X';
        } else {
            dfs(board, row, col);
        }
        
        return board;
    }
    
    private void dfs(char[][] board, int row, int col) {
        if (row < 0 || row >= board.length || col < 0 || col >= board[0].length || board[row][col] != 'E') {
            return;
        }
        
        int mines = countAdjacentMines(board, row, col);
        
        if (mines == 0) {
            board[row][col] = 'B';
            int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {1, 1}, {-1, 1}, {1, -1}};
            for (int[] dir : dirs) {
                dfs(board, row + dir[0], col + dir[1]);
            }
        } else {
            board[row][col] = (char) (mines + '0');
        }
    }
    
    private int countAdjacentMines(char[][] board, int row, int col) {
        int count = 0;
        int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}, {-1, -1}, {1, 1}, {-1, 1}, {1, -1}};
        
        for (int[] dir : dirs) {
            int newRow = row + dir[0];
            int newCol = col + dir[1];
            if (newRow >= 0 && newRow < board.length && newCol >= 0 && newCol < board[0].length && board[newRow][newCol] == 'M') {
                count++;
            }
        }
        
        return count;
    }
}

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