<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 solve(vector<vector<char>>& board) {
        
        for(int r = 0; r < board.size(); ++r){
            for(int c = 0; c < board[0].size(); ++c){
                bool isEdge = r == 0 || r == board.size() - 1 || c == 0 || c == board[0].size() - 1 ? true : false;
                
                if(dontReplace.count({r,c}) == 0 && board[r][c] == 'O' && isEdge){
                    //scan from border. any O connectedc to an O on the border should be 
                    //added to the set of dontReplace
                    //therefore any O that is not found in dontReplace is not connected to the border
                    set<pair<int,int>> visited;
                    dfs(r, c, board, visited);
                }

            }
        }

        for(int r = 0; r < board.size(); ++r){
            for(int c = 0; c < board[0].size(); ++c){
                if(board[r][c] == 'O' && dontReplace.count({r,c}) == 0) //if its an O and not found in dontReplace
                    board[r][c] = 'X';
            }
        }
    }

    void dfs(int r, int c, vector<vector<char>>& board, set<pair<int,int>>& visited){
        if(r < 0 || r == board.size() || c < 0 || c == board[0].size() || board[r][c] == 'X' || visited.count({r,c}) > 0)
            return;
        
        visited.insert({r,c});
        dontReplace.insert({r, c});

        dfs(r + 1, c, board, visited);
        dfs(r - 1, c, board, visited);
        dfs(r, c + 1, board, visited);
        dfs(r, c - 1, board, visited);
    }

private:
    set<pair<int,int>> dontReplace; 
};
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