<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 {
    void dfs(vector<vector<int>>& grid, int row, int col){
        int n = grid.size();
        int m = grid[0].size();
        if(row < 0 || row > n - 1 || col < 0 || col > m - 1 || grid[row][col] == 0) return ;

        grid[row][col] = 0;
        dfs(grid, row - 1, col);
        dfs(grid, row + 1, col) ; 
        dfs(grid, row, col - 1);
        dfs(grid, row, col + 1);
        return;
    }

public:
    int numEnclaves(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int ans = 0;
        for(int i = 0; i < n; i++){
            if(grid[i][0] == 1) dfs(grid, i, 0);
            if(grid[i][m -1] == 1) dfs(grid, i, m -1);
        }
        for(int j = 0; j < m; j++){
            if(grid[0][j] == 1) dfs(grid, 0, j);
            if(grid[n -1][j] == 1) dfs(grid, n -1, j);
        }
        for(auto V : grid)
            for(auto num : V)
                if(num) ans ++;
        return ans;
    }
};
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