<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:
    int numEnclaves(vector<vector<int>>& grid) {
            int rows = grid.size();
            int cols = grid[0].size();
            int res = 0;
            int move[5] = {-1,0,1,0,-1};
            function<void(int,int)> dfs = [&](int i, int j)
            {
                if (i < 0 || i >= rows || j < 0 || j >= cols || grid[i][j] == 0)
                    return;
                else {
                    grid[i][j] = 0;
                    res += 1;
                    for (int k = 0; k < 4; ++k)
                    {
                        dfs(i+move[k],j+move[k+1]);
                    }
                }
                
            };
            int tot = 0;
            for (int i = 0; i < rows; ++i)
            {
                for (int j = 0; j < cols; ++j)
                {
                    if (grid[i][j] == 1)
                        ++tot;
                }
            }
            for (int i = 0; i < cols; ++i)
            {
                dfs(0,i);
                dfs(rows-1,i);
            }

            for (int i = 0; i < rows; ++i)
            {
                dfs(i,0);
                dfs(i,cols-1);
            }
        return tot - res;
    }
};
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