<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:
    vector<int>topo(vector<vector<int>>& graph, vector<int>& indegree){
       vector<int>sorted;
       queue<int>q;

       for(int i=0; i<indegree.size(); i++){
        if(indegree[i]==0){
            q.push(i);
        }
       }
       while(!q.empty()){
        int node=q.front();
        q.pop();
        sorted.push_back(node);

        for(auto it:graph[node]){
            indegree[it]--;
            if(indegree[it]==0){
                q.push(it);
            }
        }
       }
       if(sorted.size()==graph.size()){
        return sorted;
       }
       return {};

    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        vector<int>ans;
        for(int i=0; i<n; i++){
            if(group[i]==-1){
               group[i]=m;
               m++;
            }
        }
        vector<vector<int>>adjlistitems(n, vector<int>());
        vector<vector<int>>adjlistgroups(m,vector<int>());
        vector<int>inditems(n,0);
        vector<int>indgrps(m,0);
        

        for(int i=0; i<n; i++){
            for(auto it:beforeItems[i]){
                adjlistitems[it].push_back(i);
                inditems[i]++;
               
            //    if(group[i]!=group[it]){
            //     adjlistgroups[group[it]].push_back(group[i]);
            //     indgrps[group[i]]++;
            //  }
            }
        }
        for(int i=0; i<n; i++){
            for(auto it:beforeItems[i]){
             if(group[i]!=group[it]){
                adjlistgroups[group[it]].push_back(group[i]);
                indgrps[group[i]]++;
             }
            }
        }
        vector<int>itemsorted=topo(adjlistitems, inditems);
        vector<int>groupsorted=topo(adjlistgroups, indgrps);

        if(itemsorted.size()==0 || groupsorted.size()==0){
            return {};
        }
       unordered_map<int,vector<int>>groupitems;

       for(auto it: itemsorted){
         groupitems[group[it]].push_back(it);
       }
       for(auto i: groupsorted){
        for(auto it: groupitems[i]){
            ans.push_back(it);
        }
       }
    //    for(int i=0; i<groupsorted.size(); i++){
    //     for(auto it: groupitems[i]){
    //         ans.push_back(it);
    //     }
    //    }
       return ans;
    }
};
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