<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:

    bool isValid(pair<int,int>p,int n,int m)
    {   
        int x = p.first;
        int y = p.second;
        return (x>=0 && y>=0 && x<n && y<m);
    }

    void bfs(pair<int,int>start,set<pair<int,int>>&visited,vector<vector<char>>& board,int n,int m,vector<int>moveX,vector<int>moveY)
    {
        queue<pair<int,int>> q;
        q.push(start);
        set<pair<int,int>> currIsland;
        bool isIsland=true;
        while(!q.empty())
        {   
            pair<int,int>curr = q.front();
            cout << curr.first << " " << curr.second << " " << isIsland << endl;
            auto it = visited.find(curr);
            q.pop();
            if(it==visited.end() && board[curr.first][curr.second]=='O')
            {
                currIsland.insert(curr);
                for(int i=0;i<4;i++)
                {
                    pair<int,int> newpair = {curr.first+moveX[i],curr.second+moveY[i]};
                    if(isValid(newpair,n,m))
                    {
                        if(board[newpair.first][newpair.second]=='O')
                        {
                            q.push(newpair);
                        }    
                    }
                    else{
                        isIsland=false;
                    }
                }
                visited.insert(curr);
            }
        }
        cout << endl;
        if(isIsland)
        {
            for(auto it:currIsland)
            {
                // cout<<"i"<<it.first<<" j"<<it.second<<endl;
                board[it.first][it.second]='X';
            }
        }
        
    }

    void solve(vector<vector<char>>& board) {
       
        int n = board.size();
        int m = board[0].size();

        vector<int> moveX = {1,0,-1,0};
        vector<int> moveY = {0,1,0,-1};
        set<pair<int,int>> visited;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {       
                auto it=visited.find({i,j});
                if(board[i][j]=='O' && it==visited.end())
                {
                    bfs({i,j},visited,board,n,m,moveX,moveY);
                }
            }
        }
    }
};
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