<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 numEnclaves(vector<vector<int>>& grid) {
        
        int n=grid.size(),m=grid[0].size();

        vector<vector<int>>vis(n,vector<int>(m,0));

        queue<pair<int,int>>q;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){

                if(grid[0][j]==1){
                    vis[0][j]=1;
                    q.push({0,j});

                }
                if(grid[n-1][j]==1){
                    vis[n-1][j]=1;
                    q.push({n-1,j});
                }
            }
        }

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){

                if(grid[i][0]==1){
                    vis[i][0]=1;
                    q.push({i,0});
                }

                if(grid[i][m-1]==1){
                    vis[i][m-1]=1;
                    q.push({i,m-1});
                }
            }
        }


        while(!q.empty()){
            int r=q.front().first;
            int c=q.front().second;
            q.pop();

            int drow[]={-1,0,1,0};
            int dcol[]={0,-1,0,1};

            for(int i=0;i<4;i++){
                int nrow=r+drow[i];
                int ncol=c+dcol[i];

                if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && !vis[nrow][ncol] && grid[nrow][ncol]==1){
                    vis[nrow][ncol]=1;
                    q.push({nrow,ncol});
                }
            }
        }

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1 && vis[i][j]==1){
                    grid[i][j]=0;
                }
            }
        }

        int cnt=0;
           for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==1 ){
                    cnt++;
                }
            }
        }

        return cnt;

    }
};
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