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

    int numEnclaves(vector<vector<int>>& grid) {
        int rows = grid.size();
        int cols = grid[0].size();
        vector<vector<int>> vis(rows, vector<int> (cols, -1));

        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(i == 0 || j == 0 || i == rows - 1 || j == cols - 1) {
                    if(grid[i][j] == 1)
                        dfs(i, j, rows, cols, grid, vis);
                }
            }
        }
        int res = 0;
        for(int i = 0; i < rows; i++) {
            for(int j = 0; j < cols; j++) {
                if(grid[i][j] == 1) res++;
            }
        }
        return res;
    }
};
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