<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 {
    private:  
    void dfs(int row, int col, vector<vector<int>> &vis, 
    vector<vector<int>> &mat) {
        vis[row][col] = 1; 
        int n = mat.size();
        int m = mat[0].size();
        vector<int>delrow = {-1, 0, +1, 0};
        vector<int>delcol = {0, 1, 0, -1}; 
        // check for top, right, bottom, left 
        for(int i = 0;i<4;i++) {
            int nrow = row + delrow[i];
            int ncol = col + delcol[i]; 
            // check for valid coordinates and unvisited Os
            if(nrow >=0 && nrow <n && ncol >= 0 && ncol < m 
            && !vis[nrow][ncol] && mat[nrow][ncol] == 1){
                dfs(nrow, ncol, vis, mat); 
            }
        }
    }
public:
    int numEnclaves(vector<vector<int>>& mat) {
        int n=mat.size();
        int m=mat[0].size();
    

        vector<vector<int>>vis(n,vector<int>(m,0));
         for(int j=0;j<m;j++){
            if(!vis[0][j]&&mat[0][j]==1){
                 dfs(0,j,vis,mat);
            }
            if(!vis[n-1][j]&&mat[n-1][j]==1){
                 dfs(n-1,j,vis,mat);
            }
         }
         for(int i=0;i<n;i++){
            if(!vis[i][0] && mat[i][0]==1){
                dfs(i,0,vis,mat);
            }
            if(!vis[i][m-1] && mat[i][m-1]==1){
                dfs(i,m-1,vis,mat);
            }
        }
        int count = 0;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (!vis[i][j] && mat[i][j] == 1)
                {
                    count++;
                }
            }
        }
        return count;
    }
};
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