<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]:
        sg = m
        for j, g in enumerate(group):
            if g == -1:
                group[j] = sg
                sg += 1

        before_groups = [set() for _ in range(sg)]
        next_groups = [set() for _ in range(sg)]
        group_items = [[] for _ in range(sg)]
        next_items = [set() for _ in range(n)]
        for i, items in enumerate(beforeItems):
            g = group[i]
            group_items[g].append(i)
            for item in items:
                next_items[item].add(i)
                bg = group[item]
                if bg != g:
                    before_groups[g].add(bg)
                    next_groups[bg].add(g)
        group_indegrees = [len(groups) for groups in before_groups]

        output = []

        def visit_items(g):
            item_indegrees = [0] * n
            for i in group_items[g]:
                for item in beforeItems[i]:
                    if item in group_items[g]:
                        item_indegrees[i] += 1 
            
            iq = [i for i in group_items[g] if item_indegrees[i] == 0]
            while iq:
                i = iq.pop()
                output.append(i)
                for nitem in next_items[i]:
                    item_indegrees[nitem] -= 1
                    if item_indegrees[nitem] == 0:
                        iq.append(nitem)

        q = [g for g, indegree in enumerate(group_indegrees) 
            if indegree == 0]

        while q:
            g = q.pop()
            visit_items(g)
            for ng in next_groups[g]:
                group_indegrees[ng] -= 1
                if group_indegrees[ng] == 0:
                    q.append(ng)
        if len(output) != n:
            return []
        return output

523