<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();
        int m=grid[0].size();
        int ans=0;
        vector<vector<int>> visited(n,vector<int>(m,0));
        for(int i=0;i<n;++i){
            for(int j=0;j<m;++j){
                if(grid[i][j]==1 && visited[i][j]==0){
                    int cnt=0;
                    queue<pair<int,int>> q;
                    q.push({i,j});
                    int onedge=0;
                    visited[i][j]=1;
                    while(!q.empty()){
                        int row=q.front().first;
                        int col=q.front().second;
                        q.pop();
                        if(row==0 || row==n-1 || col==0 || col==m-1){
                            onedge=1;
                        }
                        cnt++;
                        int rc[]={0,-1,0,1};
                        int cc[]={-1,0,1,0};
                        for(int i=0;i<4;++i){
                            int newrow=row+rc[i];
                            int newcol=col+cc[i];
                            if(newrow>=0 && newrow<n && newcol>=0 && newcol<m && visited[newrow][newcol]==0 && grid[newrow][newcol]==1){
                                visited[newrow][newcol]=1;
                                q.push({newrow,newcol});
                            }
                        }
                    }
                    if(onedge==0){
                        ans+=cnt;
                    }
                }
            }
        }
        return ans;
    }
};
52902