<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 n = grid.size();
        int m = grid[0].size();
        int count = 0;
        vector<vector<int>>vis(n,vector<int>(m,0));
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(!vis[i][j]&&grid[i][j]==1){
                    bool b = true;
                    int c = 0;
                    queue<pair<int,int>>q;
                    q.push({i,j});
                    while(!q.empty()){
                        auto s = q.front();
                        q.pop();
                        int ti = s.first, tt = s.second;
                        if (vis[ti][tt]) continue;
                        vis[ti][tt]=1;
                        if(ti==0||ti==n-1||tt==0||tt==m-1){
                            b = false;
                        }
                            c++;
                        if(ti>0&&grid[ti-1][tt]==1){
                            q.push({ti-1,tt});
                        }
                        if(ti+1<n&&grid[ti+1][tt]==1){
                            q.push({ti+1,tt});
                        }
                        if(tt>0&&grid[ti][tt-1]==1){
                            q.push({ti,tt-1});
                        }
                        if(tt+1<m&&grid[ti][tt+1]==1){
                            q.push({ti,tt+1});
                        }
                    }
                    if(b){
                        count+=c;
                    }
                }
            }
        }
        return count;
    }
};
56367