<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:
    int numEnclaves(vector<vector<int>>& grid) {
        int out=0;
        vector<pair<int,int>> dir;
        dir.push_back({-1,0});
        dir.push_back({0,-1});
        dir.push_back({1,0});
        dir.push_back({0,1});
        queue<pair<int,int>> q;
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>> vis(n,vector<int>(m,0));
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j]==1 && vis[i][j]==0){
                    int flag=0;
                    if(i==0 || j==0 || i==(n-1) || j==(m-1)){
                        flag=1;
                    }
                    q.push({i,j});
                    vis[i][j]=1;
                    int count=1;
                    while(q.size()>0){
                        int siz=q.size();
                        for(int k=0;k<siz;k++){
                            pair<int,int> pa=q.front();
                            q.pop();
                            for(int t=0;t<4;t++){
                                int nx=pa.first+dir[t].first;
                                int ny=pa.second+dir[t].second;
                                if(nx<0 || ny<0 || nx>=n || ny>=m){
                                    continue;
                                }
                                else if(nx==0 || ny==0 || nx==(n-1) || ny==(m-1)){
                                    if(grid[nx][ny]==1){
                                        flag=1;
                                        if(vis[nx][ny]==0){
                                            q.push({nx,ny});
                                            vis[nx][ny]=1;
                                        }
                                    }
                                }
                                else if(grid[nx][ny]==1 && vis[nx][ny]==0){
                                    q.push({nx,ny});
                                    vis[nx][ny]=1;
                                    count++;
                                }
                            }
                        }
                    }
                    if(flag==0){
                        out+=count;
                    }
                }
                
            }
        }
        return out;
    }
};
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