<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]:
    # assign unique id for each vertex of group '-1'
    for i in range(n):
      if group[i]==-1: group[i]=m; m+=1

    # group to vertices mapping
    grpCnt=defaultdict(list)
    for i in range(n): grpCnt[group[i]].append(i)

    # DFS for getting topo order, checking cyclic
    visited,tp,routed=set(),[],set()
    def cyclic(node,ADJ):
      routed.add(node)
      try:
        visited.add(node)
        for grp in ADJ[node]:
          if grp in routed: return True
          if grp not in visited:
            if cyclic(grp,ADJ): return True
        tp.append(node)
        return False
      finally:
        routed.remove(node)

    # construct group relations
    grpAdj=defaultdict(list)
    for i in range(n):
      for b in beforeItems[i]:
        if group[i]==group[b]: continue
        grpAdj[group[i]].append(group[b])

    # topo sort the groups
    for i in grpCnt:
      if i not in visited:
        if cyclic(i,grpAdj):
          return []

    res=[]
    # construct node relations inside groups
    vAdj=defaultdict(list)
    for i,items in enumerate(beforeItems):
      GRP=group[i]
      for b in items:
        if group[b]!=GRP: continue
        vAdj[i].append(b)

    # topo sort the vertices inside groups
    for G in tp:
      verts=grpCnt[G]
      if len(verts)==1:
        res.extend(verts)
      else:
        visited,tp,routed=set(),[],set()
        for v in verts:
          if v not in visited:
            if cyclic(v,vAdj): return []
        res.extend(tp)

    return res
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