<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 bfs(int r , int c , vector<vector<char>>&board , vector<vector<int>>&vis , vector<int>delRow , vector<int>delCol){
        int n = board.size();
        int m = board[0].size();
        vis[r][c] = 1;

        for(int i=0;i<4;i++){
            int row = r+delRow[i];
            int col = c+delCol[i];
            if(row>=0 && row<n && col>=0 && col<m && !vis[row][col] && board[row][col] == 'O'){
                bfs(row , col , board , vis , delRow , delCol);
            }
        }
    }
    void solve(vector<vector<char>>& board) {
        int n = board.size();
        int m = board[0].size();
        vector<vector<int>>vis(n,vector<int>(m,0));
        vector<int>delRow = {-1 , 0 , 1 , 0};
        vector<int>delCol = {0 , 1 , 0 , -1};

        for(int j=0;j<m;j++){
            if(!vis[0][j] && board[0][j] == 'O') bfs(0,j,board,vis,delRow,delCol);

            if(!vis[n-1][j] && board[n-1][j] == 'O') bfs(n-1,j,board,vis,delRow,delCol);
        }

        for(int i=0;i<n;i++){
            if(!vis[i][0] && board[i][0] == 'O') bfs(i,0,board,vis,delRow,delCol);

            if(!vis[i][m-1] && board[i][m-1] == 'O') bfs(i,m-1,board,vis,delRow,delCol);
        }

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(!vis[i][j] && board[i][j] == 'O') board[i][j] = 'X';
            }
        }
    }
};
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