<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 topo(vector<int>adj[],int n,vector<int>&res)
    {
        queue<int>qu;
        vector<int>ind(n,0);
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<adj[i].size();j++) ind[adj[i][j]]++;
        }
        for(int i=0;i<n;i++) if(!ind[i]) qu.push(i);
        while(!qu.empty())
        {
            int x=qu.front();qu.pop();
            res.push_back(x);
            for(int i=0;i<adj[x].size();i++) 
            {
                ind[adj[x][i]]--;
                if(ind[adj[x][i]]==0) qu.push(adj[x][i]);
            }
        }
        if(res.size()!=n) return false;
        return true;
    }
    vector<int> sortItems(int n, int m, vector<int>& group, vector<vector<int>>& beforeItems) {
        vector<int>adj[n];
        for(int i=0;i<group.size();i++) if(group[i]==-1) group[i]=m++;
        vector<int>grps[m];
        vector<int>adjg[m];
        for(int i=0;i<beforeItems.size();i++)
        {
            for(int j=0;j<beforeItems[i].size();j++)
            {
                adj[beforeItems[i][j]].push_back(i);
                if(group[i]==-1||group[beforeItems[i][j]]==-1) continue;
                auto it=find(adjg[group[beforeItems[i][j]]].begin(),adjg[group[beforeItems[i][j]]].end(),group[i]);
                if(it==adjg[group[beforeItems[i][j]]].end()&&group[beforeItems[i][j]]!=group[i])
                {
                    int gp=group[beforeItems[i][j]];
                    int gc=group[i];
                    adjg[gp].push_back(gc);
                }
            }
        }
        vector<int>topg,topi;
        bool res1=topo(adj,n,topi);
        bool res2=topo(adjg,m,topg);
        if(!res1||!res2) return {};
        vector<int>ans;
        for(int i=0;i<topi.size();i++){
            int g=group[topi[i]];
            grps[g].push_back(topi[i]);
        }
        for(int i=0;i<topg.size();i++)
        {
            for(int j=0;j<grps[topg[i]].size();j++)
            ans.push_back(grps[topg[i]][j]);
        }
        return ans;
    }
};
45962