<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 {
private:
void bfs(vector<vector<char>>& board,vector<vector<int>>& vis,int x,int y)
{
    queue<pair<int,int>> q;
    q.push({x,y});
    int row[]={-1,0,+1,0};
    int column[]={0,-1,0,+1};
    while(!q.empty())
    {
        x=q.front().first;
        y=q.front().second;
        q.pop();
        vis[x][y]=1;
    for(int i=0;i<4;i++)
    {
        int nrow=x+row[i];
       int ncolumn=y+column[i];
        if(nrow>-1 && nrow<board.size() && ncolumn>-1 && ncolumn<board[0].size())
        {
        if(vis[nrow][ncolumn]==0 && board[nrow][ncolumn]!='X')
            {
                vis[nrow][ncolumn]=1;
                q.push({nrow,ncolumn});
            }
        }
    }

    }
}
public:
    void solve(vector<vector<char>>& board) { 
        int r=board.size();
        int c=board[0].size();
        vector<vector<int>> vis(r,vector<int>(c));
        int rev=r-1;
        for(int i=0;i<c;i++)
        {
            if(board[0][i]!='X')
                vis[0][i]=1;
            if(board[r-1][i]!='X')
                vis[r-1][i]=1;
        }
        for(int i=0;i<r;i++)
        {
            if(board[i][0]!='X')
                vis[i][0]=1;
            if(board[i][c-1]!='X')
                vis[i][c-1]=1;
        }
        for(int i=0;i<c;i++)
        {
            if(board[0][i]!='X')
                bfs(board,vis,0,i);
            if(board[r-1][i]!='X')
                 bfs(board,vis,r-1,i);
        }
        for(int i=0;i<r;i++)
        {
            if(board[i][0]!='X')
                bfs(board,vis,i,0);
            if(board[i][c-1]!='X')
                bfs(board,vis,i,c-1);
        }

        for(int i=0;i<r;i++)
        {
            for(int j=0;j<c;j++)
            {
                if(vis[i][j]==0)
                    board[i][j]='X';
            }
        }
        
        }
        
    
};
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