<p>Let&#39;s play the minesweeper game (<a href="https://en.wikipedia.org/wiki/Minesweeper_(video_game)" target="_blank">Wikipedia</a>, <a href="http://minesweeperonline.com" target="_blank">online game</a>)!</p>

<p>You are given an <code>m x n</code> char matrix <code>board</code> representing the game board where:</p>

<ul>
	<li><code>&#39;M&#39;</code> represents an unrevealed mine,</li>
	<li><code>&#39;E&#39;</code> represents an unrevealed empty square,</li>
	<li><code>&#39;B&#39;</code> represents a revealed blank square that has no adjacent mines (i.e., above, below, left, right, and all 4 diagonals),</li>
	<li>digit (<code>&#39;1&#39;</code> to <code>&#39;8&#39;</code>) represents how many mines are adjacent to this revealed square, and</li>
	<li><code>&#39;X&#39;</code> represents a revealed mine.</li>
</ul>

<p>You are also given an integer array <code>click</code> where <code>click = [click<sub>r</sub>, click<sub>c</sub>]</code> represents the next click position among all the unrevealed squares (<code>&#39;M&#39;</code> or <code>&#39;E&#39;</code>).</p>

<p>Return <em>the board after revealing this position according to the following rules</em>:</p>

<ol>
	<li>If a mine <code>&#39;M&#39;</code> is revealed, then the game is over. You should change it to <code>&#39;X&#39;</code>.</li>
	<li>If an empty square <code>&#39;E&#39;</code> with no adjacent mines is revealed, then change it to a revealed blank <code>&#39;B&#39;</code> and all of its adjacent unrevealed squares should be revealed recursively.</li>
	<li>If an empty square <code>&#39;E&#39;</code> with at least one adjacent mine is revealed, then change it to a digit (<code>&#39;1&#39;</code> to <code>&#39;8&#39;</code>) representing the number of adjacent mines.</li>
	<li>Return the board when no more squares will be revealed.</li>
</ol>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img src="https://assets.leetcode.com/uploads/2023/08/09/untitled.jpeg" style="width: 500px; max-width: 400px; height: 269px;" />
<pre>
<strong>Input:</strong> board = [[&quot;E&quot;,&quot;E&quot;,&quot;E&quot;,&quot;E&quot;,&quot;E&quot;],[&quot;E&quot;,&quot;E&quot;,&quot;M&quot;,&quot;E&quot;,&quot;E&quot;],[&quot;E&quot;,&quot;E&quot;,&quot;E&quot;,&quot;E&quot;,&quot;E&quot;],[&quot;E&quot;,&quot;E&quot;,&quot;E&quot;,&quot;E&quot;,&quot;E&quot;]], click = [3,0]
<strong>Output:</strong> [[&quot;B&quot;,&quot;1&quot;,&quot;E&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;1&quot;,&quot;M&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img src="https://assets.leetcode.com/uploads/2023/08/09/untitled-2.jpeg" style="width: 489px; max-width: 400px; height: 269px;" />
<pre>
<strong>Input:</strong> board = [[&quot;B&quot;,&quot;1&quot;,&quot;E&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;1&quot;,&quot;M&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;]], click = [1,2]
<strong>Output:</strong> [[&quot;B&quot;,&quot;1&quot;,&quot;E&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;1&quot;,&quot;X&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;B&quot;],[&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;,&quot;B&quot;]]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 50</code></li>
	<li><code>board[i][j]</code> is either <code>&#39;M&#39;</code>, <code>&#39;E&#39;</code>, <code>&#39;B&#39;</code>, or a digit from <code>&#39;1&#39;</code> to <code>&#39;8&#39;</code>.</li>
	<li><code>click.length == 2</code></li>
	<li><code>0 &lt;= click<sub>r</sub> &lt; m</code></li>
	<li><code>0 &lt;= click<sub>c</sub> &lt; n</code></li>
	<li><code>board[click<sub>r</sub>][click<sub>c</sub>]</code> is either <code>&#39;M&#39;</code> or <code>&#39;E&#39;</code>.</li>
</ul>

class Solution {
public:
    bool inboard(const vector<vector<char>>& board, int x, int y) {
        return (x >= 0 && x < board.size() && y >= 0 && y < board[0].size());
    }
    void dfs(vector<vector<char>>& board, int x, int y) {
        vector<vector<int>> dirs = {{1,-1},{1,0},{1,1},{0,-1},{0,1},{-1,-1},{-1,0},{-1,1}};
        // Search 8 directions for mines
        int count = 0;
        for(auto d : dirs) {
            int nx = x + d[0];
            int ny = y + d[1];
            if(inboard(board, nx, ny) && board[nx][ny] == 'M') {
                count++;
            }
        }
        // Rule 2: If at least one mine is present, place their count.
        if(count > 0) {
            board[x][y] = count + '0';
        } 
        // Rule 3: If no mine is present, mark current as 'B' and explore neighbors.
        else {
            board[x][y] = 'B';
            for(auto d : dirs) {
                int nx = x + d[0];
                int ny = y + d[1];
                if(inboard(board, nx, ny) && board[nx][ny] == 'E') {
                    // Recursive DFS call for empty square
                    board[nx][ny] = 'B'; // Mark as visited before the call
                    dfs(board, nx, ny);
                }
            }
        }
    }
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        int r = click[0];
        int c = click[1];
        // Rule 1: If a mine 'M' is revealed, change it to 'X' (game over).
        if(board[r][c] == 'M') {
            board[r][c] = 'X';
        } 
        else {
            dfs(board, r, c);
        }
        return board;
    }
};

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