<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 dfs(int row, int col, vector<vector<int>> &vis, vector<vector<int>> &mat, int delrow[], int delcol[]) {
        vis[row][col] = 1; 
        int n = mat.size();
        int m = mat[0].size();
        
        // 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, delrow, delcol); 
            }
        }
    }

    int fill(int n, int m,vector<vector<int>> mat)
    {
        int delrow[] = {-1, 0, +1, 0};
        int delcol[] = {0, 1, 0, -1}; 
        vector<vector<int>> vis(n, vector<int>(m,0)); 
        // traverse first row and last row 
        for(int j = 0 ; j<m;j++) {
            // check for unvisited Os in the boundary rows
            // first row 
            if(!vis[0][j] && mat[0][j] == 1) {
                dfs(0, j, vis, mat, delrow, delcol); 
            }
            
            // last row 
            if(!vis[n-1][j] && mat[n-1][j] == 1) {
                dfs(n-1,j,vis,mat, delrow, delcol); 
            }
        }
        
        for(int i = 0;i<n;i++) {
            // check for unvisited Os in the boundary columns
            // first column 
            if(!vis[i][0] && mat[i][0] == 1) {
                dfs(i, 0, vis, mat, delrow, delcol); 
            }
            
            // last column
            if(!vis[i][m-1] && mat[i][m-1] == 1) {
                dfs(i, m-1, vis, mat, delrow, delcol); 
            }
        }
        
        // if unvisited O then convert to X
        int cnt=0;
        for(int i = 0;i<n;i++) {
            for(int j= 0 ;j<m;j++) {
                if(!vis[i][j] && mat[i][j] == 1) 
                    cnt++;
            }
        }
        
        return cnt; 
    }
    int numEnclaves(vector<vector<int>>& grid) {

        int n = grid.size();
        int m = grid[0].size();

       return  fill(n,m,grid);
        
    }
};
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