<p>You are given an <code>m x n</code> binary matrix <code>grid</code>, where <code>0</code> represents a sea cell and <code>1</code> represents a land cell.</p>

<p>A <strong>move</strong> consists of walking from one land cell to another adjacent (<strong>4-directionally</strong>) land cell or walking off the boundary of the <code>grid</code>.</p>

<p>Return <em>the number of land cells in</em> <code>grid</code> <em>for which we cannot walk off the boundary of the grid in any number of <strong>moves</strong></em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves1.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,0],[1,0,1,0],[0,1,1,0],[0,0,0,0]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> There are three 1s that are enclosed by 0s, and one 1 that is not enclosed because its on the boundary.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/02/18/enclaves2.jpg" style="width: 333px; height: 333px;" />
<pre>
<strong>Input:</strong> grid = [[0,1,1,0],[0,0,1,0],[0,0,1,0],[0,0,0,0]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All 1s are either on the boundary or can reach the boundary.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 500</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
</ul>

class Solution {
public:

    int BFS(vector<vector<int>>&grid, vector<vector<int>>&visited, int sr, int sc, int n,int m)
    {
        queue<pair<int,int>>q;
        int count=1;
        q.push(pair<int,int>(sr,sc));
        visited[sr][sc]=1;
        int flag = 1;
        while(!q.empty())
        {
            pair<int,int>p = q.front();
            q.pop();

            int x = p.first;
            int y = p.second;

            if(x-1>=0)
            {
                if(grid[x-1][y]==1 && visited[x-1][y]==0)
                {
                    count++;
                    q.push(pair<int,int>(x-1,y));
                }
                visited[x-1][y]=1;
            }

            if(x+1<n)
            {
                if(grid[x+1][y]==1 && visited[x+1][y]==0)
                {
                    count++;
                    q.push(pair<int,int>(x+1,y));
                }
                visited[x+1][y]=1;

            }

            if(y-1>=0)
            {
                if(grid[x][y-1]==1 && visited[x][y-1]==0)
                {
                    count++;
                    q.push(pair<int,int>(x,y-1));
                }
                visited[x][y-1]=1;

            }


            if(y+1<m)
            {
                if(grid[x][y+1]==1 && visited[x][y+1]==0)
                {
                    count++;
                    q.push(pair<int,int>(x,y+1));
                }
                visited[x][y+1]=1;
            }
        }

        return count;
    }
    int numEnclaves(vector<vector<int>>& grid) {

        vector<vector<int>>visited;

        int n=grid.size();
        int m = grid[0].size();
        int totalcount=0;
        int i,j;
        for(i=0;i<n;i++)
        {
            vector<int> vrow;
            for(j=0;j<m;j++)
            {
                if(grid[i][j]==1)
                    totalcount++;
                vrow.push_back(0);
            }
            
            visited.push_back(vrow);
        }

        //int ans=0;

        for(i=0;i<n;i++)
        {
           
            if(grid[i][0]==1 && visited[i][0]==0)
            {
                totalcount -=BFS(grid,visited,i,0,n,m);
            }

            if(grid[i][m-1]==1 && visited[i][m-1]==0)
            {
                totalcount -=BFS(grid,visited,i,m-1,n,m);
            }
            
        }
        
        for(j=0;j<m;j++)
        {
           
            if(grid[0][j]==1 && visited[0][j]==0)
            {
                totalcount-=BFS(grid,visited,0,j,n,m);
            }

            if(grid[n-1][j]==1 && visited[n-1][j]==0)
            {
                totalcount-=BFS(grid,visited,n-1,j,n,m);
            }
            
        }

        return totalcount;
        
    }
};
40197