<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:
    vector<vector<bool>> vis;
    vector<pair<int, int>> dir{{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
    int numEnclaves(vector<vector<int>>& grid) {
        int cnt = 0;
        int m = grid.size();
        int n = grid[0].size();
        vis.resize(m);
        for (int i = 0; i < m; i++) {
            vis[i].resize(n, false);
        }
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (i == 0 || i == m - 1 || j == 0 || j == n - 1) {
                    if (grid[i][j] == 1 && !vis[i][j]) {
                        dfs(grid, {i, j}, cnt);
                    }
                }
            }
        }
        int total = 0;

        for (int i = 0; i < m; i++)
            for (int j = 0; j < n; j++)
                if (grid[i][j] == 1)
                    total++;
        cout << total << " " << cnt << endl;
        return total - cnt;
    }

    void dfs(vector<vector<int>>& v, pair<int, int> p, int& cnt) {

        int a = p.first;
        int b = p.second;
        vis[a][b] = true;
        cnt++;
        for (int i = 0; i < 4; i++) {
            int x = a + dir[i].first;
            int y = b + dir[i].second;

            if (x >= 0 && y >= 0 && x < v.size() && y < v[0].size() &&
                !vis[x][y] && v[x][y] == 1)
                dfs(v, {x, y}, cnt);
        }
    }
};
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