Minesweeper
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0529.Minesweeper/images/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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0500-0599/0529.Minesweeper/images/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>
Depth-First Search; Breadth-First Search; Array; Matrix
class Solution {
public:
    vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
        int m = board.size(), n = board[0].size();
        int i = click[0], j = click[1];

        function<void(int, int)> dfs = [&](int i, int j) {
            int cnt = 0;
            for (int x = i - 1; x <= i + 1; ++x) {
                for (int y = j - 1; y <= j + 1; ++y) {
                    if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M') {
                        ++cnt;
                    }
                }
            }
            if (cnt) {
                board[i][j] = cnt + '0';
            } else {
                board[i][j] = 'B';
                for (int x = i - 1; x <= i + 1; ++x) {
                    for (int y = j - 1; y <= j + 1; ++y) {
                        if (x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'E') {
                            dfs(x, y);
                        }
                    }
                }
            }
        };

        if (board[i][j] == 'M') {
            board[i][j] = 'X';
        } else {
            dfs(i, j);
        }
        return board;
    }
};