<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:

   bool topo(vector<int> &ans,vector<int> mp[],vector<int> &in){
      queue<int> q;
      for(int i=0;i<in.size();i++){
        if(in[i]==0){
              q.push(i);
        }
      }

      while(!q.empty()){
        int u= q.front(); q.pop();
        ans.push_back(u);

        for(int v:mp[u]){
            in[v]--;
            if(in[v]==0) q.push(v);
        }
      }

      return (ans.size()==in.size());
    
   }
    vector<int> sortItems(int n, int m, vector<int>& grp, vector<vector<int>>& prev) {
    for(int i=0;i<n;i++){
        if(grp[i]==-1) grp[i]=m++;
    }

    vector<int> nd(n,0), gd(m,0);
    vector<int> mpn[n], mpg[m];

    for(int v=0;v<n;v++){
        for(int u:prev[v]){
            nd[v]++;
            mpn[u].push_back(v); // nodes ki directed list

            if(grp[u]!=grp[v]){
                gd[grp[v]]++;  //grps ki directed list
                mpg[grp[u]].push_back(grp[v]);
            }
        }
    }
     vector<int> qitm, qgrp; // dono ki toposorted fixed sequence
    bool itm= topo(qitm, mpn,nd);
    bool grps= topo(qgrp,mpg,gd);

    if(!itm || !grps) return {};

      vector<int> ans;
      map<int,vector<int>> mp;
      for(int it:qitm){
         mp[grp[it]].push_back(it);
      }
      
      for(int x: qgrp){   
          for(auto y:mp[x]){ // jo grp pehle aaya ho uske elements bhar do jo ki pehle toposorted way me aaye hai 
            ans.push_back(y);
          }
      }

      return ans;

    }
};
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