<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:

    bool isValid(int n,int m,int i,int j){
        return i<n && i>=0 && j<m && j>=0;
    }

    int bfs(int i,int j,vector <vector<int>> &grid,vector <vector<int>> &visited,bool &boundry){ // bfs will be more suitable here
        queue <pair<int,int>> q;
        int n = grid.size(),m = grid[0].size();
        q.push({i,j});
        if(i==0 || i==n-1 || j==0 || j==m-1){
            boundry = 0;
        }
        visited[i][j] = 1;

        int dx[4] = {1,-1,0,0};
        int dy[4] = {0,0,-1,1};
        int count = 1;
        while(!q.empty()){
            auto [x,y] = q.front();
            q.pop();
            for(int I=0;I<4;I++){
                int newI = x + dx[I];
                int newJ = y + dy[I];
                if(isValid(n,m,newI,newJ) && visited[newI][newJ]==-1 && grid[newI][newJ]==1){
                    count++;
                    visited[newI][newJ] = 1;
                    // grid[newI][newJ] = 0;
                    q.push({newI,newJ});
                    if(newI==0 || newI==n-1 || newJ==0 || newJ==m-1){
                        boundry = 0;
                    }
                }
            }
        }

        return count;
    }

    int numEnclaves(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        vector<vector <int>> visited(n,vector <int> (m,-1));

        int ans = 0;

        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(visited[i][j]==-1 && grid[i][j]){
                    bool boundry = 1;
                    cout<<i<<' '<<j<<' '<<boundry<<' ';
                    int count = bfs(i,j,grid,visited,boundry);
                    ans = ans + boundry * count;
                }   
            }
        }

        return ans;
    }
};
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