<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:
    vector<pair<int,int>>dir = {{0,1},{0,-1},{1,0},{-1,0}} ;
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size() , n = grid[0].size() ;
        vector<vector<bool>> vis(m , vector<bool>(n , 0)) ;
        for(int i = 0 ; i<n ; i++){
            if(grid[0][i]==1 && !vis[0][i]){
                dfs(0 , i , vis , grid) ;
            }
        }
        for(int i = 0 ; i<n ; i++){
            if(grid[m-1][i]==1 && !vis[m-1][i]){
                dfs(m-1 , i , vis , grid) ;
            }
        }
        for(int i = 0 ; i<m ; i++){
            if(grid[i][0]==1 && !vis[i][0]){
                dfs(i , 0 , vis , grid) ;
            }
        }
        for(int i = 0 ; i<m ; i++){
            if(grid[i][n-1]==1 && !vis[i][n-1]){
                dfs(i , n-1 , vis , grid) ;
            }
        }
        int ans = 0 ;
        for(int i = 0 ; i <m ; i++){
            for(int j = 0 ; j<n ; j++){
                if(grid[i][j]==1 && !vis[i][j]){
                    ans++ ;
                }
            }
        }
        return ans ;
    }
private:
    void dfs(int x , int y , vector<vector<bool>> &vis , vector<vector<int>> &grid){
        if(vis[x][y]) return ;
        vis[x][y]=1 ;
        for(auto [r,c]:dir){
            int newx = x+r ;
            int newy = y+c ;
            if(newx==grid.size() || newx<0 || newy==grid[0].size() || newy<0 || grid[newx][newy]==0 || vis[newx][newy]) continue ;
            dfs(newx , newy , vis , grid) ;
        }
    }
};
48667