<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>

#include <vector>
#include <queue>

using namespace std;

class Solution {
public:
    int numEnclaves(vector<vector<int>>& grid) {
        queue<pair<int,int>> q; 
        int n = grid.size(); 
        int m = grid[0].size(); 
        vector<vector<int>> vis(n, vector<int>(m, 0)); // Use a vector for vis
        
        // Traverse boundary elements
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                // Check first row, first col, last row, last col 
                if(i == 0 || j == 0 || i == n-1 || j == m-1) {
                    // If it is land, store it in queue
                    if(grid[i][j] == 1) {
                        q.push({i, j}); 
                        vis[i][j] = 1; 
                    }
                }
            }
        }
        
        int delrow[] = {-1, 0, +1, 0};
        int delcol[] = {0, +1, +0, -1}; 
        
        // BFS to mark all reachable land cells from the boundary
        while(!q.empty()) {
            int row = q.front().first; 
            int col = q.front().second; 
            q.pop(); 
            
            // Traverse all 4 directions
            for(int i = 0; i < 4; i++) {
                int nrow = row + delrow[i];
                int ncol = col + delcol[i]; 
                // Check for valid coordinates and for land cell
                if(nrow >= 0 && nrow < n && ncol >= 0 && ncol < m 
                && vis[nrow][ncol] == 0 && grid[nrow][ncol] == 1) {
                    q.push({nrow, ncol});
                    vis[nrow][ncol] = 1; 
                }
            }
        }
        
        int cnt = 0;
        // Count unvisited land cells (enclaves)
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < m; j++) {
                // Check for unvisited land cell
                if(grid[i][j] == 1 && vis[i][j] == 0) 
                    cnt++; 
            }
        }
        return cnt; 
    }
};

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