<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>

using namespace std;

class Solution {
private:
    vector<int> delRow = {0, -1, 0, 1};
    vector<int> delCol = {-1, 0, 1, 0};

    // Helper function for DFS
    void bfs(int row, int col, vector<vector<int>>& grid, vector<vector<bool>>& visited, queue<pair<int, int>>& q) {
        int m = grid.size();
        int n = grid[0].size();
        
        q.push({row,col});
        visited[row][col]=1;

        while(!q.empty()){

            for (int d = 0; d < 4; ++d) {
                int newRow = q.front().first + delRow[d];
                int newCol = q.front().second + delCol[d];

                if (newRow >= 0 && newRow < m && newCol >= 0 && newCol < n &&
                    grid[newRow][newCol] == 1 && !visited[newRow][newCol]) {
                    q.push({newRow,newCol});
                    visited[newRow][newCol]=1;
                }
            }
            q.pop();
        }
    }

public:
    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
         queue<pair<int, int>> q;
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        
        for (int i = 0; i < m; ++i) {
            if (grid[i][0] == 1 && !visited[i][0]) bfs(i, 0, grid, visited,q);
            if (grid[i][n - 1] == 1 && !visited[i][n - 1]) bfs(i, n - 1, grid, visited, q);
        }
        
        for (int j = 0; j < n; ++j) {
            if (grid[0][j] == 1 && !visited[0][j]) bfs(0, j, grid, visited, q);
            if (grid[m - 1][j] == 1 && !visited[m - 1][j]) bfs(m - 1, j, grid, visited, q);
        }
    
        int enclaves = 0;
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1 && !visited[i][j]) {
                    enclaves++;
                }
            }
        }
        
        return enclaves;
    }
};

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