<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:
typedef pair<int, int> P;
    P find(P i, map<P, P>& parent){
        if( parent[i] == i){
            return i;
        }

        return parent[i] = find(parent[i], parent);
    }

    void Union(P a, P b, map<P, P>& parent, int& n, int& m){
        P a_parent = find(a, parent);
        P b_parent = find(b, parent);

        int b_row = b_parent.first;
        int b_col = b_parent.second;

        if(b_row == 0 || b_row == n-1 || b_col == 0 || b_col == m-1){
            parent[a_parent] = b_parent;
        }
        else{
            parent[b_parent] = a_parent;
        }
    }
    void solve(vector<vector<char>>& board) {
        map<P, P> parent;
        int n = board.size();
        int m = board[0].size();

        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                parent[{i,j}] = {i, j};
            }
        }

        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(board[i][j] == 'O'){
                    //up
                    if(i-1 >=0 && board[i-1][j] == 'O'){
                        Union({i, j}, {i-1, j}, parent, n, m);
                    }

                    //down
                    if(i+1 < n && board[i+1][j] == 'O'){
                        Union({i, j}, {i+1, j}, parent, n, m);
                    }

                    //left
                    if(j-1 >=0 && board[i][j-1] == 'O'){
                        Union({i, j}, {i, j-1}, parent, n, m);
                    }

                    //right
                    if(j+1 <m && board[i][j+1] == 'O'){
                        Union({i, j}, {i, j+1}, parent, n, m);
                    }
                }
            }
        }

        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(board[i][j] == 'O'){
                    P c_parent = find({i,j}, parent);
                    int row = c_parent.first;
                    int col = c_parent.second;
                    if(row != 0 && row != n-1 && col != 0 && col != m-1){
                        board[i][j] = 'X';
                    }
                }
            }
        }
    }
};
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