<p>There are&nbsp;<code>n</code>&nbsp;items each&nbsp;belonging to zero or one of&nbsp;<code>m</code>&nbsp;groups where <code>group[i]</code>&nbsp;is the group that the <code>i</code>-th item belongs to and it&#39;s equal to <code>-1</code>&nbsp;if the <code>i</code>-th item belongs to no group. The items and the groups are zero indexed. A group can have no item belonging to it.</p>

<p>Return a sorted list of the items such that:</p>

<ul>
	<li>The items that belong to the same group are next to each other in the sorted list.</li>
	<li>There are some&nbsp;relations&nbsp;between these items where&nbsp;<code>beforeItems[i]</code>&nbsp;is a list containing all the items that should come before the&nbsp;<code>i</code>-th item in the sorted array (to the left of the&nbsp;<code>i</code>-th item).</li>
</ul>

<p>Return any solution if there is more than one solution and return an <strong>empty list</strong>&nbsp;if there is no solution.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2019/09/11/1359_ex1.png" style="width: 191px; height: 181px;" /></strong></p>

<pre>
<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
<strong>Output:</strong> [6,3,4,1,5,2,0,7]
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
<strong>Output:</strong> []
<strong>Explanation:</strong>&nbsp;This is the same as example 1 except that 4 needs to be before 6 in the sorted list.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= m &lt;= n &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>group.length == beforeItems.length == n</code></li>
	<li><code>-1 &lt;= group[i] &lt;= m - 1</code></li>
	<li><code>0 &lt;= beforeItems[i].length &lt;= n - 1</code></li>
	<li><code>0 &lt;= beforeItems[i][j] &lt;= n - 1</code></li>
	<li><code>i != beforeItems[i][j]</code></li>
	<li><code>beforeItems[i]&nbsp;</code>does not contain&nbsp;duplicates elements.</li>
</ul>

class Solution {
public:
void f(vector<int> adj[],vector<int> &in,vector<int> &v){
    queue<int> q;
    int n=in.size();
    for(int i=0;i<n;i++){
        if(in[i]==0){
            q.push(i);
        }
    }
    while(!q.empty()){
        int node=q.front();
        q.pop();
        v.push_back(node);
        for(auto it:adj[node]){
            in[it]--;
            if(in[it]==0){
                q.push(it);
            }
        }
    }
}
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& bi) {
       vector<int> ans,gtopo,ntopo;
    //    return ans;
    //    vector<unordered_set<int>> v(m);
       vector<int> grpdeg(m+n,0),indegree(n,0),nadj[n],grpadj[m+n];
       for(int i=0;i<n;i++){
        if(group[i]==-1){
            group[i]=m++;
        }
       }
        for(int i=0;i<n;i++){
            for(int j=0;j<bi[i].size();j++){
                int dal=group[bi[i][j]];
                int node=bi[i][j];
                nadj[node].push_back(i);
                indegree[i]++;
                if(dal==group[i]){
                    continue;
                }
                
                grpadj[dal].push_back(group[i]);
                grpdeg[group[i]]++;
            }
        }
        f(nadj,indegree,ntopo);
        f(grpadj,grpdeg,gtopo);
        if(ntopo.size()!=n ){
            return ans;
        }
        vector<int> sampl[m+n+1];
        for(auto it:ntopo){
            int dal=group[it];
            sampl[dal].push_back(it);
        }
        for(auto it:gtopo){
            for(auto itt:sampl[it]){
                ans.push_back(itt);
            }
        }
        if(ans.size()!=n){
            return {};
        }
        return ans;   
    }
};
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