<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 dx[4] = {-1,1,0,0};
int dy[4] = {0,0,1,-1};

void dfs(int x,int y,vector<vector<bool> > &vis,vector<vector<int>>& grid)
{
    vis[x][y] = 1;

    for(int k=0;k<4;k++)
    {
        int nx = x+dx[k];
        int ny = y+dy[k];

        if(nx>=0 && nx<grid.size() && ny>=0 && ny<grid[0].size() && vis[nx][ny]==0 && grid[nx][ny]==1)
        {
            dfs(nx,ny,vis,grid);
        }
    }
}

    int numEnclaves(vector<vector<int>>& grid) 
    {
        set<pair<int,int> > edges;
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<bool> > visited(n,vector<bool>(m,false));

        int counttotal = 0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(grid[i][j]==1)
                {
                    counttotal+=1;
                    if(i==0 || i==n-1 || j==0 || j==m-1)
                    {
                        edges.insert({i,j});
                    }
                }
            }
        }

        //dfs
        for(auto &it : edges)
        {
            dfs(it.first,it.second,visited,grid);
        }

        int marked = 0;
          for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                if(visited[i][j]==1)
                {
                   marked++;
                }
            }
        }
        return counttotal-marked;
    }
};
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