<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> toposort(int n,vector<int> adj[], vector<int> &indegree){
        vector<int> ans;
        queue<int> q;
        for(int i=0; i<n; i++){
            if(indegree[i]==0){
                q.push(i);
            }
        }
        while(!q.empty()){
            int node=q.front();
            q.pop();
            ans.push_back(node);
            for(auto it:adj[node]){
                indegree[it]--;
                if(indegree[it]==0){
                    q.push(it);
                }
            }
        }
        return ans;
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int tot_grp=m;
        for(int i=0; i<n; i++){
            if(group[i]==-1){
                group[i]=tot_grp;
                tot_grp++;
            }
        }
        vector<int> adj1[n];
        vector<int> adj2[tot_grp];
        vector<int> indegree1(n,0);
        vector<int> indegree2(tot_grp,0);
        for(int i=0; i<n; i++){
            for(auto j:beforeItems[i]){
                adj1[j].push_back(i);
                indegree1[i]++;
                if(group[j]!=group[i]){
                    indegree2[group[i]]++;
                    adj2[group[j]].push_back(group[i]);
                }
            }
        }
        vector<int> topo1=toposort(n,adj1,indegree1);
        vector<int> topo2=toposort(tot_grp,adj2,indegree2);
        if(topo1.size()!=n || topo2.size()!=tot_grp){
            return {};
        }
        vector<int> ans;
        vector<vector<int>> temp;
        map<int,vector<int>> mp;
        for(auto it:topo1){
            mp[group[it]].push_back(it);
        }
        for(auto it:topo2){
            temp.push_back(mp[it]);
        }
        for(auto it=temp.begin();it!=temp.end();it++){
            for(auto j:*it){
                ans.push_back(j);
            }
        }
        return ans;
    }
};
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