<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(unordered_map<int,list<int>>&adj,vector<int>&indegree){
        queue<int>q;
        for(int i=0;i<indegree.size();i++){
            if(indegree[i]==0)q.push(i);
        }
        vector<int>ans;
         while(!q.empty()){
             int front=q.front();
             ans.push_back(front);
             q.pop();
             for(auto nbr: adj[front]){
                  indegree[nbr]--;
                  if(indegree[nbr]==0)q.push(nbr);

             }
         }
      if (ans.size() != indegree.size()) return {}; // return empty if cycle exists
        
        return ans;
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        unordered_map<int,list<int>>listadj;
        unordered_map<int,list<int>>groupadj;


        for (int i = 0; i < n; i++) {
            if (group[i] == -1) group[i] = m++;
        }

        vector<int>degreelist(n,0);
        vector<int>degreegroup(m,0);
        for(int i=0;i<group.size();i++){ 
            for(auto v: beforeItems[i]){
                listadj[v].push_back(i);
                degreelist[i]++;

                    if (group[v] != group[i]) {  // Only add group dependency if they are different
                    groupadj[group[v]].push_back(group[i]);
                    degreegroup[group[i]]++;
                }
            }

        }
                    vector<int>topolist=topo(listadj,degreelist);
            vector<int>topogroup=topo(groupadj,degreegroup);
            unordered_map<int,list<int>>tempans;
                    for(auto v: topolist){
                tempans[group[v]].push_back(v);
            }

            vector<int>ans;
          for(auto v: topogroup){
                ans.insert(ans.end(),tempans[v].begin(),tempans[v].end());
            }

            return ans;
    }
};
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