<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 topoSortUtil(vector<int>& topoOrder, unordered_map<int,int>& indeg, unordered_map<int,vector<int>>& graph){
        priority_queue<int,vector<int>, greater<int>> minHeap;
        for(auto &i: indeg) if(!i.second) minHeap.push(i.first);
        int cnt=0;
        while(!minHeap.empty()){
            int tp=minHeap.top(); minHeap.pop();
            topoOrder.push_back(tp);
            cnt++;
            for(auto &u: graph[tp]){
                if(!--indeg[u]) minHeap.push(u);
            }
        }
        return cnt == graph.size();
    }

    bool topoSort(vector<int>& topoOrder, unordered_map<int,vector<int>>& graph){
        int n=graph.size();
        unordered_map<int,int> indeg;
        for(auto &i: graph){
            if(!indeg.count(i.first)) indeg[i.first]=0;
            for(auto &j: i.second) indeg[j]++;
        }
        return topoSortUtil(topoOrder,indeg,graph);
    }

    vector<int> sortItems(int n, int m, vector<int>& g, vector<vector<int>>& dep) {
        unordered_map<int,unordered_map<int,vector<int>>> groupAndItsGraph;
        for(int i=0;i<n;++i){
            if(g[i]==-1) g[i]=m++;
            groupAndItsGraph[g[i]][i]={};
        }
        unordered_map<int,vector<int>> groupGraph;
        for(int i=0;i<m;++i) groupGraph[i]={};
        for(int i=0;i<n;++i){
            for(auto &j: dep[i]){
                if(g[i]==g[j]) groupAndItsGraph[g[j]][j].push_back(i);
                else groupGraph[g[j]].push_back(g[i]);
            }
        }
        vector<int> groupTopoOrder, nodesTopoOrder;
        if(!topoSort(groupTopoOrder, groupGraph)) return {};
        for(auto &i: groupTopoOrder) if(!topoSort(nodesTopoOrder, groupAndItsGraph[i])) return {};
        return nodesTopoOrder;
    }   
};
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