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
func updateBoard(board [][]byte, click []int) [][]byte {
	m, n := len(board), len(board[0])
	i, j := click[0], click[1]

	var dfs func(i, j int)
	dfs = func(i, j int) {
		cnt := 0
		for x := i - 1; x <= i+1; x++ {
			for y := j - 1; y <= j+1; y++ {
				if x >= 0 && x < m && y >= 0 && y < n && board[x][y] == 'M' {
					cnt++
				}
			}
		}
		if cnt > 0 {
			board[i][j] = byte(cnt + '0')
			return
		}
		board[i][j] = 'B'
		for x := i - 1; x <= i+1; x++ {
			for 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
}