<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>


from collections import defaultdict
from typing import List


class Solution:
    def sortItems(
        self, n: int, m: int, group: List[int], beforeItems: List[List[int]]
    ) -> List[int]:
        graph = [[] for _ in range(m)]
        for i in range(n):
            if group[i] < 0:
                group[i] = m
                m += 1
                graph.append([])
            graph[group[i]].append(i)

        groups = defaultdict(set)
        adj = [defaultdict(set) for _ in range(m)]
        for i, lst in enumerate(beforeItems):
            ig = group[i]
            for j in lst:
                jg = group[j]
                if ig == jg:
                    adj[ig][i].add(j)
                else:
                    groups[ig].add(jg)
            if not lst:
                adj[ig][i]

        def toposort(nodelst, adj: dict):
            ret = []
            ind = {i: 0 for i in nodelst}
            for v in adj.values():
                for i in v:
                    ind[i] += 1
            q = [i for i, v in ind.items() if v == 0]

            while q:
                x = q.pop()
                ret.append(x)
                for i in adj[x]:
                    ind[i] -= 1
                    if ind[i] == 0:
                        q.append(i)
            return ret if len(ret) == len(nodelst) else None

        gs = toposort(range(m), groups)
        if gs is None:
            return []
        ret = []
        for i in gs:
            ns = toposort(graph[i], adj[i])
            if ns is None:
                return []
            ret.extend(ns)
        return ret[::-1]

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