<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) {
       if (grid.empty() || grid[0].empty()) return 0;
        
        int m = grid.size();
        int n = grid[0].size();
        vector<vector<int>> directions = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        
        
        auto bfs = [&](int x, int y) {
            queue<pair<int, int>> q;
            q.push({x, y});
            grid[x][y] = 2; 
            while (!q.empty()) {
                auto [cx, cy] = q.front();
                q.pop();
                
                for (const auto& dir : directions) {
                    int nx = cx + dir[0];
                    int ny = cy + dir[1];
                    
                    if (nx >= 0 && nx < m && ny >= 0 && ny < n && grid[nx][ny] == 1) {
                        grid[nx][ny] = 2; 
                        q.push({nx, ny});
                    }
                }
            }
        };
        
        for (int i = 0; i < m; ++i) {
            if (grid[i][0] == 1) bfs(i, 0);
            if (grid[i][n - 1] == 1) bfs(i, n - 1);
        }
        
        for (int j = 0; j < n; ++j) {
            if (grid[0][j] == 1) bfs(0, j);
            if (grid[m - 1][j] == 1) bfs(m - 1, j);
        }
       
        int count = 0;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    ++count;
                }
            }
        }
        
        return count; 
    }
};
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