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

        
        function<void(int,int)>dfs=[&](int i,int j){
            visited[i][j]=1;

            int dx[]={0,1,-1,0};
            int dy[]={1,0,0,-1};

            for(int k=0;k<4;k++){
                int nr=dx[k]+i,nc=dy[k]+j;
                if(nr>0 and nr<n-1 and nc>0 and nc<m-1 and adj[nr][nc]==1 and visited[nr][nc]==0){
                    dfs(nr,nc);
                }
            }
        };

        for(int i=0;i<n;i++){
            if(adj[i][0])dfs(i,0);
            if(adj[i][m-1])dfs(i,m-1);
        }
        for(int j=0;j<m;j++){
            if(adj[0][j])dfs(0,j);
            if(adj[n-1][j])dfs(n-1,j);
        }
        int ans=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(adj[i][j]==1 and visited[i][j]==0)ans++;
            }
        }
        return ans;
    }
};
45202