<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> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        int groupId = m;
        map<int, set<int>> groupItem;

        for(int i=0;i<n;i++){
            if(group[i]==-1){
                group[i]=(groupId++);
            }
            groupItem[group[i]].insert(i);
        }

        map<int, set<int>> next;
        map<int, int> indegree;

        for(int i=0;i<n;i++){
            for(int j: beforeItems[i]){
                if(group[i]!=group[j]){continue;}
                next[j].insert(i);
                indegree[i]++;
            }
        }

        map<int, vector<int>> groupOrdered;
        for(int i=0;i<groupId;i++){
            if(groupItem[i].size()==0){continue;}
            groupOrdered[i] = TS(groupItem[i], next, indegree);

            // cout<<"group:"<<i<<endl;
            // for(int num: groupOrdered[i]){cout<<num<<' ';}
            // cout<<endl;
            if(groupOrdered[i].size()==0){return {};}
        }

        next.clear();
        indegree.clear();

        for(int i=0;i<n;i++){
            for(int j: beforeItems[i]){
                if(group[i]==group[j]){continue;}
                if(next[group[j]].find(group[i])==next[group[j]].end())
                {
                    next[group[j]].insert(group[i]);
                    indegree[group[i]]++;
                }
                
            }
        }

        set<int> allItem;
        for(int i=0;i<groupId;i++){allItem.insert(i);}
        // for(int i: allItem){cout<<i<<' ';}cout<<endl;
        vector<int> order = TS(allItem, next, indegree);
        // cout<<"order:"<<order.size()<<endl;
        if(order.size()==0){return {};}

        vector<int> res;
        for(int i: order){
            for(int item: groupOrdered[i]){
                res.push_back(item);
            }
        }
        return res;
    }

    vector<int> TS(set<int>& groups, map<int, set<int>> &next, map<int, int>& indegree){
        queue<int> q;
        vector<int> res;
        for(int node: groups){
            if(indegree[node]==0){
                q.push(node);
            }
        }

        while(!q.empty()){
            int cur = q.front();
            q.pop();
            res.push_back(cur);

            for(int next: next[cur]){
                indegree[next]--;
                if(indegree[next]==0){
                    q.push(next);
                }
            }
        }
        
        if(res.size()==groups.size()){return res;}
        return {};
    }
};
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