<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>>& grid, vector<vector<int>>& visited){

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

        visited[row][col] = 1;

        int rowdir[] = {-1,1,0,0};
        int coldir[] = {0,0,-1,1};

        for(int i=0;i<4;i++){
            int nrow = row + rowdir[i];
            int ncol = col + coldir[i];

            if(nrow>=0 && nrow<n && ncol>=0 && ncol<m && grid[nrow][ncol]==1 && !visited[nrow][ncol]){

                dfs(nrow,ncol,grid,visited);
            }
        }
    }
public:
    int numEnclaves(vector<vector<int>>& grid) {
        
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<int>> visited(n, vector<int>(m,0));

        for(int i=0;i<m;i++){
            if(grid[0][i] == 1){
                dfs(0,i,grid,visited);
            }
        }

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

        for(int i=0;i<n;i++){
            if(grid[i][0] == 1){
                dfs(i,0,grid,visited);
            }
        }

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

        int cnt = 0;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j] == 1 && !visited[i][j])cnt++;
            }
        }

        return cnt;

    }
};
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