<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(int i,int j,vector<vector<int>>& grid){
        int m = grid.size(), n = grid[0].size();
        grid[i][j]=0;
        vector<pair<int,int>> dir = {{-1,0},{1,0},{0,-1},{0,1}};
        for(auto [dx,dy]:dir){
            int nr = i+dx, nc = j+dy;
            if(nr>=0 && nr<m && nc>=0 && nc<n && grid[nr][nc]==1){
                dfs(nr,nc,grid);
            }
        }
    }
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        for(int i=0; i<m; i++){
            if(grid[i][0]==1){
                dfs(i,0,grid);
            }
            if(grid[i][n-1]==1){
                dfs(i,n-1,grid);
            }
        }
        for(int i=0; i<n; i++){
            if(grid[0][i]==1){
                dfs(0,i,grid);
            }
            if(grid[m-1][i]==1){
                dfs(m-1,i,grid);
            }
        }
        int count = 0;
        for(int i=0; i<m; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j]==1) count++;
            }
        }
        return count;
    }
};
52517