<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 sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        for i in range(n):
            if group[i]==-1:
                group[i]=i+m

        graph0={}
        seen0=[0]*(m+n)
        graph1={}
        seen1=[0]*n

        for i,x in enumerate(beforeItems):
            for j in x:
                if group[j]!=group[i]:
                    graph0.setdefault(group[j],[]).append(group[i])
                    seen0[group[i]]+=1

                graph1.setdefault(j,[]).append(i)
                seen1[i]+=1

        def fn(graph,seen):
            N=len(seen)
            ans=[]
            stack=[k for k in range(N) if seen[k]==0 ]
            while stack:
                n=stack.pop()
                ans.append(n)
                for s in graph.get(n,[]):
                    seen[s]-=1
                    if seen[s]==0:
                        stack.append(s)
            return ans

        top0=fn(graph0,seen0)
        a=len(top0)
        b=len(seen0)       
        if a!=b:
            return []

        top1=fn(graph1,seen1)
        c=len(top1)
        d=len(seen1)
        if c!=d:
            return []
        map0={x:i for i,x in enumerate(top0)}
        map1={x:i for i,x in enumerate(top1)}                                     
        return sorted( range(n),key=lambda x:(map0[group[x]],map1[x]))
356