<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:
    def top_sort(self,adj,nodes):
        res=[]
        path=set()
        visit=set()
        def dfs(node):
            if node in visit:
                return True
            visit.add(node)
            path.add(node)
            for nei in adj[node]:
                if nei in path:
                    return False
                if not dfs(nei):
                    return False
            res.append(node)
            path.remove(node)
            return True
    
        for node in nodes:
            if not dfs(node):
                return []

        return res[::-1]



    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        group_to_items=defaultdict(list)
        pseudo_group=m
        for i,g in enumerate(group):
            if g==-1:
                group_to_items[pseudo_group].append(i)
                group[i]=pseudo_group
                pseudo_group+=1
            else:
                group_to_items[g].append(i)
        
        adj_g=defaultdict(set)
        adj=defaultdict(list)
        
        for i,pres in enumerate(beforeItems):
            for pre in pres:
                gi=group[i]
                gp=group[pre]
                if gi!=gp:
                    adj_g[gp].add(gi)
                else:
                    adj[pre].append(i)
        
        
        top_sorted=defaultdict(list)
        for g,nodes in group_to_items.items():
            s=self.top_sort(adj,nodes)
            if s==[]:
                return []
            else:
                top_sorted[g]=s
        
        top_sorted_groups=self.top_sort(adj_g,list(range(pseudo_group)))
        if top_sorted_groups==[]:
            return []
        res=[]
        for g in top_sorted_groups:
            res+=top_sorted[g]
        
        return res
         
    
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