<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) {
    unordered_map<int, set<int>> groupGraph;
    unordered_map<int, vector<int>> groupAndItems;
    unordered_map<int, int> itemAndGroup;
    int idxNewGroup = m;
    for(int i = 0; i < group.size(); i++) {
      if(group[i] == -1) {
        groupAndItems[idxNewGroup].push_back(i);
        itemAndGroup[i] = idxNewGroup++;
      } else {
        groupAndItems[group[i]].push_back(i);
        itemAndGroup[i] = group[i];
      } 
    }

    for(auto &[group, items]: groupAndItems) {
      unordered_map<int, set<int>> itemsGraph;
      for(int &item: items) {
        vector<int> before = beforeItems[item];
        for(int &num: before) {
          if(itemAndGroup[num] != group) continue;
          itemsGraph[num].insert(item);
        }
      }
      for(int &item: items) 
        if(!itemsGraph.count(item)) itemsGraph[item] = {};
      items = apply_khan(itemsGraph);
      if(items.size() == 0) return {};
      for(int &item: items) {
        vector<int> before = beforeItems[item];
        for(int &num: before) {
          int before_group = itemAndGroup[num];
          if(before_group == group) continue;
          groupGraph[itemAndGroup[num]].insert(group);
        }
      }
    }

    for(auto &[group, items]: groupAndItems) { 
      if(!groupGraph.count(group)) groupGraph[group] = {};
    }

    vector<int> result;
    vector<int> x = apply_khan(groupGraph);
    for(int &group: x) {
      result.insert(result.end(), groupAndItems[group].begin(), groupAndItems[group].end());
    }
    return result;
  }

  vector<int> apply_khan(unordered_map<int, set<int>> graph) {
    unordered_map<int, int> in_going;
    for(auto& [from, to]: graph) {
      if(!in_going.count(from)) in_going[from] = 0;
      for(int num: to) in_going[num]++;
    }

    vector<int> result;
    queue<int> q;
    for(auto &[x, y]: in_going)
      if(!y) q.push(x);

    while(q.size()) {
      int current = q.front(); q.pop();
      result.push_back(current);
      for(int child: graph[current]) {
        in_going[child]--;
        if(!in_going[child]) q.push(child);
      }
    }

    if(result.size() != in_going.size()) return {};
    return result;
  }
};
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