<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 dir[4][2]={{-1,0},{1,0},{0,-1},{0,1}};
    void dfs(vector<vector<int>>& g, int i, int j){
        g[i][j]=0;
        for(int k=0;k<4;++k){
            int x=i+dir[k][0], y=j+dir[k][1];
            if(x>=0 and y>=0 and x<g.size() and y<g[0].size() and g[x][y]==1){
                dfs(g,x,y);
            }
        }
    }
    int numEnclaves(vector<vector<int>>& g) {
        int n=g.size(), m=g[0].size();
        for(int i=0;i<n;++i){
            for(int j=0;j<m;j++){
                if(i==0 or j==0 or i==n-1 or j==m-1){
                    if(g[i][j]==1){
                        dfs(g,i,j);
                    }
                }
            }
        }
        int ans=0;
        for(int i=0;i<n;++i){
            for(int j=0;j<m;++j){
                if(g[i][j]==1) ans++;
            }
        }
        return ans;
    }
};
30957