<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 , int m , int n , int i , int j, vector<vector<int>>&vis)
    {
        int count = 0 ;
        int flag = 0 ;

        int xdel[] = {0,1,0,-1};
        int ydel[] = {-1,0,1,0};

        queue<pair<int,int>>q;
        vector<pair<int,int>>pairs;

        vis[i][j]= 1;
        if(i == 0 || i== m-1 || j == 0 || j == n-1)
        {
            flag = 1;
        }
        q.push({i,j});
        pairs.push_back({i,j});
        count ++;

        while(!q.empty())
        {
            int x = q.front().first ;
            int y = q.front().second;

            q.pop();

            for( int i = 0 ;i < 4;i++)
            {
                int xtemp = x + xdel[i];
                int ytemp = y + ydel[i];

                if(xtemp >=0 && xtemp <m && ytemp>=0 && ytemp<n && grid[xtemp][ytemp]==1 && (!vis[xtemp][ytemp]))
                {
                    vis[xtemp][ytemp]=1;

                    if(xtemp == 0 || xtemp == m-1 || ytemp == 0 || ytemp == n-1)
                    {
                        flag =1 ;
                    }
                    count++;
                    q.push({xtemp,ytemp});
                }
            }
        }
        
        if(flag == 0) return count ;
        return 0 ;
        
    }
    int numEnclaves(vector<vector<int>>& grid) {

        int m = grid.size();
        int n = grid[0].size();

        vector<vector<int>>vis(m,vector<int>(n,0));
        int count = 0;
        
        for(int i =0;i<m;i++)
        {
            for(int j =0;j<n;j++)
            {
                if(!vis[i][j] && grid[i][j]==1)
                {
                    count +=bfs(grid,m,n,i,j,vis);
                }
                else if(vis[i][j]==0)
                {
                    vis[i][j]=1;
                }
            }
        }

        return count ;
        
    }
};
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