<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]:
            
        # put everything into groups, edges between groups, top sort, top sort again
        graph = defaultdict(list) # node -> to edges
        groups = defaultdict(list) # group -> list of nodes
        groupEdges = defaultdict(set) # group -> to edges
        
        id_ = m
        for i in range(n):
            if group[i] == -1:
                group[i] = id_
                id_ += 1

        groupIndegrees = { i: 0 for i in range(0, id_) }
        indegrees = { i: 0 for i in range(0, n) }
        for i in range(n):
            groups[group[i]].append(i)
            for from_ in beforeItems[i]:
                if group[i] not in groupEdges[group[from_]] and group[i] != group[from_]:
                    groupEdges[group[from_]].add(group[i])
                    groupIndegrees[group[i]] += 1
                graph[from_].append(i)
                indegrees[i] += 1

        orderedGroups = []
        def groupTopSort(node):
            orderedGroups.append(node)
            for cgroup in groupEdges[node]:
                groupIndegrees[cgroup] -= 1
                if groupIndegrees[cgroup] == 0:
                    groupTopSort(cgroup)

        for gr, fr in list(groupIndegrees.items()): # deep copy
            if fr == 0:
                groupTopSort(gr)

        if len(orderedGroups) != id_:
            return []

        orderedNodes = []
        def topSort(node):
            orderedNodes.append(node)
            for child in graph[node]:
                indegrees[child] -= 1
                if indegrees[child] == 0:
                    topSort(child)

        for gr, fr in list(indegrees.items()): # deep copy
            if fr == 0:
                topSort(gr)

        if len(orderedNodes) != n:
            return []

        ordering = defaultdict(int)

        for index, i in enumerate(orderedNodes):
            ordering[i] = index
        
        for gr in groups:
            groups[gr].sort(key = lambda x: ordering[x])
        
        ans = []
        return [v for i in orderedGroups for v in groups[i]]
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