<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]:
        size = n # == len(beforeItems)

        # topological sort
        def topsort(item, need, visited, output):
            # recursive implementation

            # each item only once
            if item in visited:
                return
            visited.add(item)

            # visit prerequisites
            for next in need[item]:
                topsort(next, need, visited, output)

            # found the proper position
            output.append(item)

        # wrapper for topsort
        def mysort(need):
            output  = []
            visited = set()

            if isinstance(need, list):
                keys = range(len(need)) # list
            else:
                keys = need.keys()      # dict

            for node in keys:
                topsort(node, need, visited, output)

            # consistency check (loops are not allowed)
            have = set()
            for o in output:
                for n in need[o]:
                    if n not in have:
                        return [] # cyclic dependency found
                have.add(o)

            return output

        # convert "no group" to unique group ids
        limit = max(group) + 1
        for item in range(size):
            if group[item] == -1:
                group[item] = limit
                limit      += 1

        # slice items into groups (I call them "blocks")
        blocks = [ [] for _ in range(limit) ]
        for item in range(size):
            blocks[group[item]].append(item)

        # order of blocks
        order = [ [] for _ in range(limit) ] # I tried defaultdict(set) first, 
                                             # but lists are even faster for the given input
                                             # despite their O(n) lookup

        # sort each block
        for id in range(limit):
            if not blocks[id]:
                continue # unused group ID

            # extract prerequisites that happen to be in the same block
            need = { s: [] for s in blocks[id] }
            for s in blocks[id]:
                for b in beforeItems[s]:
                    if b in blocks[id]:
                        # depends on an item in the same block
                        need [s] .append(b)
                    else:
                        # depends on an item in a different block
                        order[id].append(group[b])

            blocks[id] = mysort(need)
            # abort if cycle was found
            if not blocks[id]:
                return []

        # sort blocks
        output = mysort(order)

        # push, push, push !
        for o in output:
            if blocks[o]:
                yield from blocks[o]
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