<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 dx[4] = {1, -1, 0, 0};
    int dy[4] = {0, 0, 1, -1};
    bool isvalid(int a, int b, int x, int y) {
        return (a >= 0 && a < x && b >= 0 && b < y);
    }
    int numEnclaves(vector<vector<int>>& grid) {
        int landcells = 0;
        int rows = grid.size();
        int cols = grid[0].size();

        queue<pair<int, int>> q;
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                landcells += grid[i][j];
                if (i * j == 0 || i == rows - 1 || j == cols - 1)
                    q.push({i, j});
            }
        }
        vector<vector<bool>> vis(rows, vector<bool>(cols, false));
        while (!q.empty()) {
            auto [x, y] = q.front();
            q.pop();
            if (!grid[x][y]||vis[x][y])
                continue;
            for (int i = 0; i < 4; i++) {
                int nx = x + dx[i], ny = y + dy[i];
                if (!isvalid(nx, ny, rows, cols)||!grid[nx][ny]||vis[nx][ny])
                    continue;
                q.push({nx, ny});
            }
            vis[x][y] = 1;
            landcells--;
        }
        return landcells;
    }
};
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