<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 m ,n;

    void dfs(vector<vector<int>>& grid , vector<vector<bool>>&visited , int i , int j) {

        if(i<0 || i>=m || j<0 || j>=n || visited[i][j] || grid[i][j]==0 ) return ;

        visited[i][j] = true;
        dfs(grid,visited,i+1,j);
        dfs(grid,visited,i-1,j);
        dfs(grid,visited,i,j+1);
        dfs(grid,visited,i,j-1);

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

        vector<vector<bool>>visited(m,vector<bool>(n,false));
        
        for(int i = 0 ; i<m ; i++) {
            for(int j = 0 ; j<n ; j++) {
                if((grid[i][j]==1) && (visited[i][j]==false) &&(i==0 || i==m-1 || j==0 || j==n-1) ) {
                    dfs(grid , visited , i , j);
                }
            }
        }
        
        int count = 0 ;
        for(int i = 0 ; i<m ; i++) {
            for(int j = 0 ; j<n ; j++) {
                if(visited[i][j]==false && grid[i][j]==1) 
                count++;
            }
        } 
        
        return count;

    }
};
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