<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:

 void dfs(vector<vector<int>>& grid,int i,int j,int &n,int&m)
 {
     if(grid[i][j]==0)return;
     grid[i][j]=0;

     vector<int>vt1={0,0,1,-1},vt2={-1,1,0,0};

     for(int k=0;k<4;k++)
     {
           int i1=i+vt1[k];
           int j1=j+vt2[k];

           if(i1>-1 && i1<n && j1>-1 && j1<m)
           {  if(grid[i1][j1])
              dfs(grid,i1,j1,n,m);
           }

     }

 }

    int numEnclaves(vector<vector<int>>& grid) 
    {
        int n=grid.size();
        int m=grid[0].size();
       for(int i=0;i<n;i++)
       { if(grid[i][0])
        dfs(grid,i,0,n,m);
        //dfs(grid,0,i,n,m);
        if(grid[i][m-1])
        dfs(grid,i,m-1,n,m);
        //dfs(grid,n-1,i,n,m);

       } 

       for(int i=0;i<m;i++)
       {
        if(grid[0][i])
        //dfs(grid,i,0,n,m);
        dfs(grid,0,i,n,m);
        if(grid[n-1][i])
        //dfs(grid,i,m-1,n,m);
        dfs(grid,n-1,i,n,m);

       } 
       
       int count=0;

       for(auto it:grid)
       {
        for(auto itt:it)
        {
            if(itt==1)count++;
        }
       }

return count;

    }
};
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