<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,int>inner_group;
 unordered_map<int,vector<int>>outter_group;
vector<int> inner_item(n);
vector<vector<int>> outter_item(n);
unordered_map<int,vector<int>> all;
        for(int i=0;i<n;i++){
            for(auto&num:beforeItems[i]){
                 if(group[i] != group[num]){
                      int g1 = (group[i] ==-1)?-i-1:group[i];
                      int g2 = (group[num] ==-1)?-num-1:group[num];
                      inner_group[g1]++;
                      outter_group[g2].push_back(g1);
                      
                  }
                  else {
                      if(group[i]==-1){
                          int g1 = -i-1;
                          int g2 = -num-1;
                          inner_group[g1]++;
                          outter_group[g2].push_back(g1);
                      }
                      else{
                          inner_item[i]++;
                          outter_item[num].push_back(i);
                      }
                  }
                  
             }
             int g = (group[i] ==-1)?-i-1:group[i];
             all[g].push_back(i);
            
        }
        vector<int> ans;
        ans.reserve(n);
        queue<int> group_q;
        for(auto&v:all){
            if(inner_group[v.first]==0)group_q.push(v.first);
        }
        while(!group_q.empty()){
            int curr_group=group_q.front();
            group_q.pop();
            queue<int> item_q;
            for(auto&num:all[curr_group]){
                if(inner_item[num]==0)item_q.push(num);
            }
            while(!item_q.empty()){
                int curr_item = item_q.front();
                item_q.pop();
                ans.push_back(curr_item);
                for(auto&num:outter_item[curr_item]){
                    inner_item[num]--;
                    if(inner_item[num]==0){
                        item_q.push(num);
                    }
                }
                
            }
            for(auto&g:outter_group[curr_group]){
                inner_group[g]--;
                if(inner_group[g]==0){
                     group_q.push(g);
                }
            }
  
        }
        return ans.size() == n ? ans : vector<int>{};
        
    }
};

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