<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:
    int flag=0;
    vector<int> fin_ans;
    int d1[100010]={0};
    int d2[100010]={0};
    void toposort(map<int,vector<int>> & grp,int g,vector<vector<int>> &adj_e){
        queue<int> q;
        for(auto x:grp[g]){
            if(d1[x]==0) q.push(x);
        }
        while(q.size()){
            auto x=q.front();q.pop();
            fin_ans.push_back(x);
            for(auto y:adj_e[x]){
                d1[y]--;
                if(d1[y]==0) q.push(y);
            }
        }
        for(auto x:grp[g]){
            if(d1[x]!=0) flag=1;
        }
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int x=m;
        map<int,vector<int>> grp;
        for(int i=0;i<n;i++){
            if(group[i]==-1) {group[i]=x;x++;}
            grp[group[i]].push_back(i);
        }
        
        vector<vector<int>> adj_e(n);
        vector<vector<int>> adj_g(x);
        //for(int i=0;i<m-1;i++) {adj_g[i].push_back(i+1);d2[i+1]++;}
        for(int i=0;i<n;i++){
            for(auto to:beforeItems[i]){
                if(group[i]==group[to]){
                    adj_e[to].push_back(i);
                    d1[i]++;
                }
                else{
                    adj_g[group[to]].push_back(group[i]);
                    d2[group[i]]++;
                }
            }
        }

        queue<int> q;
        vector<int> temp_ans;
        for(int i=0;i<x;i++) if(d2[i]==0) q.push(i);
        while(q.size()){
            int g=q.front();q.pop();
            //cout<<g<<endl;
            toposort(grp,g,adj_e);
            if(flag) return temp_ans;
            for(auto y:adj_g[g]){
                d2[y]--;
                if(d2[y]==0) q.push(y);
            }

        }
        for(int i=0;i<x;i++) if(d2[i]!=0){return temp_ans;}
        return fin_ans;
    }
};
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