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

        vector<vector<bool>> vis(m, vector<bool>(n, false));

        auto bfs = [&](int r, int c) {
            queue<pair<int, int>> q;
            q.push({r, c});
            vis[r][c] = true;
            while (!q.empty()) {
                int x = q.front().first, y = q.front().second;
                q.pop();
                int row[4] = {-1, 1, 0, 0}, col[4] = {0, 0, -1, 1};
                for (int k = 0; k < 4; k++) {
                    int ar = x + row[k], ac = y + col[k];
                    if (ar>=0 && ar<m && ac>=0 && ac<n && grid[ar][ac]==1 && !vis[ar][ac]){
                        vis[ar][ac] = true;
                        q.push({ar, ac});
                    }
                }
            }
        };

        for (int j = 0; j < n; j++) {
            if (grid[0][j] == 1 && !vis[0][j]) bfs(0, j);
            if (grid[m - 1][j] == 1 && !vis[m - 1][j]) bfs(m - 1, j);
        }
        for (int i = 0; i < m; i++) {
            if (grid[i][0] == 1 && !vis[i][0]) bfs(i, 0);
            if (grid[i][n - 1] == 1 && !vis[i][n - 1]) bfs(i, n - 1);
        }

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

        return rl;
    }
};

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