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

        int no1=0;


        //put 2 from where we can go out

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(i==0 || j==0 || i==m-1 || j==n-1){
                    if(grid[i][j]) grid[i][j]=2;
                }
            }
        }

        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if( i==0 || j==0 || i==m-1 || j==n-1 ){
                    if(grid[i][j]==2){
                        dfs(grid,i,j);
                    }
                    
                }
            }
        }


        for(int i=0;i<m;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    no1++;
                }
            }
        }
        return no1;


    }

    void dfs(vector<vector<int>>& grid ,int x,int y){
        vector<pair<int, int>> dirs = {{1, 0},{-1, 0},{0, -1},{0, 1}};
        for (auto [dx, dy] : dirs) {
        int X = x + dx;
        int Y = y + dy;
        if(X>0 && Y>=0 && X<grid.size() && Y<grid[0].size() && grid[X][Y]==1){
            grid[X][Y]=2;
            dfs(grid,X,Y);
        }
    }
    }
};
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