<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 bfs(vector<vector<int>>& board, vector<vector<bool>> &vis, int &si, int &sj, int &cells, bool &edge ){
        queue<pair<int, int>> q;
        q.push({si, sj});
        cells=1;
        vis[si][sj]=1;
        int n = board.size(), m = board[0].size();
        int drow[] = { 0, 1,  0, -1 };
        int dcol[] = { 1, 0, -1,  0 };
        while( !q.empty() ){
            int i = q.front().first;
            int j = q.front().second;
            q.pop();
            if( i==0 || j==0 || i==n-1 || j==m-1 )edge=1;
            for(int k=0; k<4; k++){
                int ni=i+drow[k];
                int nj=j+dcol[k];
                if(ni>=0 && nj>=0 && ni<n && nj<m && board[ni][nj]==1 && vis[ni][nj]==0){
                    q.push({ni, nj});
                    cells++;
                    vis[ni][nj]=1;
                }
            }
        }
    }
    void dfs( vector<vector<int>>& grid, vector<vector<bool>> &vis, int &si, int &sj, int &cells , bool &edge){
        int n = grid.size(), m = grid[0].size();
        stack<pair<int, int>> s;
        s.push( { si, sj } );
        while( !s.empty() ){
            int i = s.top().first;
            int j = s.top().second;
            s.pop();
            if( i==0 || j==0 || i==n-1 || j==m-1 ) edge = true ;
            // cells = cells + 1;
            cells++;
            vis[i][j] = 1;
            int drow[] = {1, 0, -1,  0};
            int dcol[] = {0, 1,  0, -1};
            for(int k=0; k<4; k++){
                int nrow = i + drow[k] , ncol = j + dcol[k];
                if( nrow>=0 && ncol>=0 && nrow<n && ncol<m && grid[nrow][ncol]==1 && vis[nrow][ncol]==0 )
                   s.push( { nrow , ncol } );
            }
        }
    }
    int numEnclaves(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        int ans = 0;
        vector<vector<bool>> vis(n, vector<bool>(m, 0));
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                bool has_edge = 0;
                int cells = 0 ;
                if( grid[i][j] == 1  &&  vis[i][j] == 0 ) bfs( grid, vis, i, j, cells, has_edge );
                if( has_edge == 0 )ans += cells;
            }
        }
    return ans;    
    }
};
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