<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:
//similar to surrounded region question
//doing by bfs
    vector<vector<int>> dir = {{0,1}, {0,-1}, {1,0} , {-1,0}};

    int numEnclaves(vector<vector<int>>& grid) {
        int m = grid.size(), n = grid[0].size();
        vector<vector<bool>> visited(m, vector<bool> (n, false));
        queue<pair<int,int>> q;

        //mark all the land boundary cells and the cells connected to it as visited
        //after marking them visited we will count all the non visited land cells
        //and that will be our ans

        //checking the upper boundary row and lower boundary row
        for(int j = 0; j<n; j++){
            if(grid[0][j] == 1){
                q.push({0,j});
                visited[0][j] = true;
            }

            if(grid[m-1][j] == 1){
                q.push({m-1, j});
                visited[m-1][j] = true;
            }
        }

        //checing the left boundary column and right boundary column
        for(int i = 0; i < m; i++){
            if(grid[i][0] == 1){
                q.push({i, 0});
                visited[i][0] = true;
            }

            if(grid[i][n-1] == 1){
                q.push({i, n-1});
                visited[i][n-1] = true;
            }
        }

        while(!q.empty()){
            pair<int,int> coord = q.front();
            int i = coord.first;
            int j = coord.second;
            q.pop();

            for(auto d : dir){
                int new_i = i + d[0];
                int new_j = j + d[1];

                if(new_i >= 0 && new_i < m && new_j >= 0 && new_j < n ){
                    if(grid[new_i][new_j] == 1 && !visited[new_i][new_j]){
                        visited[new_i][new_j] = true;
                        q.push({new_i, new_j});
                    }
                }
            }
        }

        int count = 0;
        for(int i = 0; i<m; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1 && !visited[i][j])
                    count++;
            }
        }

        return count;
    }
};
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