<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 bfs(int row, int col, vector<vector<int>>& grid, vector<vector<int>>& vis) {
        int n = grid.size();
        int m = grid[0].size();
        if(row == 0 || row == n - 1 || col == 0 || col == m - 1){
            return 0;
        }
        int drow[] = {-1, 0, 1, 0};
        int dcol[] = {0, 1, 0, -1};
        queue<pair<int, int>> q;
        q.push({row, col});
        vis[row][col] = 1;
        int cnt = 1; // Start with 1 to include the starting cell
        bool isEnclave = true; // Assume it's an enclave unless proven otherwise
        
        while (!q.empty()) {
            int r = q.front().first;
            int c = q.front().second;
            q.pop();
            
            // Check adjacent cells
            for (int i = 0; i < 4; i++) {
                int nrow = r + drow[i];
                int ncol = c + dcol[i];
                
                // If the new cell is within bounds and is a land cell
                if (nrow >= 0 && nrow < n && ncol >= 0 && ncol < m && grid[nrow][ncol] == 1 && vis[nrow][ncol] == 0) {
                    vis[nrow][ncol] = 1;
                    q.push({nrow, ncol});
                    cnt++;

                    // If we reach a boundary, it's not an enclave
                    if (nrow == 0 || nrow == n - 1 || ncol == 0 || ncol == m - 1) {
                        isEnclave = false;
                    }
                }
            }
        }
        
        // Return the count only if it's a valid enclave
        return isEnclave ? cnt : 0;
    }

    int numEnclaves(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int ans = 0;
        vector<vector<int>> vis(n, vector<int>(m, 0));
        
        // Traverse the entire grid
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                // Start BFS if it's an unvisited land cell
                if (grid[i][j] == 1 && vis[i][j] == 0) {
                    ans += bfs(i, j, grid, vis);
                    cout<<"ans; "<<ans<<endl;
                }
            }
        }
        return ans;
    }
};

53287