<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:
void bfs(int row,int col,  vector<vector<int>>&vis,vector<vector<int>>& grid){
    int n=grid.size();
    int m=grid[0].size();
    vis[row][col]=1;
    queue<pair<int,int>>q;
    q.push({row,col});
    vector<pair<int,int>>dir={{0,1},{1,0},{0,-1},{-1,0}};
    while(q.size()>0){
        int a=q.front().first;
        int b=q.front().second;
        q.pop();
        for(auto it :dir){
            int c=a+it.first;
            int d=b+it.second;
            if(c>=0 && c<n && d>=0 && d<m && grid[c][d]==1&& vis[c][d]==0){
                vis[c][d]=1;
                q.push({c,d});
            }
        }
    }
    return ;
}
    int numEnclaves(vector<vector<int>>& grid) {
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>vis(n,vector<int>(m,0));
        for(int i=0; i<m; i++){
            if(grid[0][i]==1){
                bfs(0,i,vis,grid);
            }
            if(grid[n-1][i]==1){
                bfs(n-1,i,vis,grid);
            }
        }
        for(int i=0; i<n; i++){
            if(grid[i][0]==1){
                bfs(i,0,vis,grid);
            }
            if(grid[i][m-1]==1){
                bfs(i,m-1,vis,grid);
            }
        }
        int count=0;
        for(int i=0; i<n; i++){
            for(int j=0; j<m; j++){
                if(grid[i][j]==1 && vis[i][j]==0){
                    count++;
                }
            }
        }
        return count;
    }
};
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