<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>>& v) {
         ios_base::sync_with_stdio(0), cin.tie(0);
 int n=v.size(),m=v[0].size();

vector<vector<int>> vis(n,vector<int>(m,0));
vector<int> row={-1,1,0,0},col={0,0,1,-1};
int cnt=0;
 function<void(int r,int c)>DFS=[&](int r,int c)->void{
    vis[r][c]=1;
    cnt++;
    for(int k=0;k<4;k++)
    {
        int nrow=r+row[k];
        int ncol=c+col[k];
        if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && !vis[nrow][ncol]&& v[nrow][ncol]==1)
        DFS(nrow,ncol);
    }
 };
 int tot=0;
 for(int i=0;i<n;i++){
    for(int j=0;j<m;j++){
        if(v[i][j]==1)
        tot++;
        if(v[i][j]==1 && (i==0||i==(n-1)|| j==0|| j==(m-1)) && !vis[i][j])
        DFS(i,j);
    }
 }
 return (tot-cnt);
    }
};
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