<p>You are given an <code>m x n</code> matrix <code>board</code> containing <strong>letters</strong> <code>&#39;X&#39;</code> and <code>&#39;O&#39;</code>, <strong>capture regions</strong> that are <strong>surrounded</strong>:</p>

<ul>
	<li><strong>Connect</strong>: A cell is connected to adjacent cells horizontally or vertically.</li>
	<li><strong>Region</strong>: To form a region <strong>connect every</strong> <code>&#39;O&#39;</code> cell.</li>
	<li><strong>Surround</strong>: The region is surrounded with <code>&#39;X&#39;</code> cells if you can <strong>connect the region </strong>with <code>&#39;X&#39;</code> cells and none of the region cells are on the edge of the <code>board</code>.</li>
</ul>

<p>A <strong>surrounded region is captured</strong> by replacing all <code>&#39;O&#39;</code>s with <code>&#39;X&#39;</code>s in the input matrix <code>board</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;O&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;X&quot;,&quot;X&quot;,&quot;X&quot;],[&quot;X&quot;,&quot;O&quot;,&quot;X&quot;,&quot;X&quot;]]</span></p>

<p><strong>Explanation:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/19/xogrid.jpg" style="width: 367px; height: 158px;" />
<p>In the above diagram, the bottom region is not captured because it is on the edge of the board and cannot be surrounded.</p>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[&quot;X&quot;]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[[&quot;X&quot;]]</span></p>
</div>

<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;= 200</code></li>
	<li><code>board[i][j]</code> is <code>&#39;X&#39;</code> or <code>&#39;O&#39;</code>.</li>
</ul>

class Solution {
public:
    void dfs(vector<vector<char>>& A, int i, int j, int m, int n) {
        if(i < 0 || i >= m || j < 0 || j >= n || A[i][j] != 'O') return;
        A[i][j] = '#'; // Temporarily mark the 'O' cells connected to the boundary
        vector<int> a = {1, -1, 0, 0}, b = {0, 0, 1, -1};
        for(int p = 0; p < 4; p++) {
            dfs(A, i + a[p], j + b[p], m, n);
        }
    }

    void solve(vector<vector<char>>& A) {
        int m = A.size(), n = A[0].size();
        if(m <= 2 || n <= 2) return;
        
        // Traverse the boundary and mark connected 'O's
        for(int i = 0; i < m; i++) {
            if(A[i][0] == 'O') dfs(A, i, 0, m, n);
            if(A[i][n-1] == 'O') dfs(A, i, n-1, m, n);
        }
        for(int j = 0; j < n; j++) {
            if(A[0][j] == 'O') dfs(A, 0, j, m, n);
            if(A[m-1][j] == 'O') dfs(A, m-1, j, m, n);
        }
        
        // Convert all remaining 'O' to 'X' and revert '#' back to 'O'
        for(int i = 0; i < m; i++) {
            for(int j = 0; j < n; j++) {
                if(A[i][j] == 'O') A[i][j] = 'X';
                else if(A[i][j] == '#') A[i][j] = 'O';
            }
        }
    }
};

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