<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 Counter, deque


class Graph:
    def __init__(self, num_nodes: int):
        self.adj = [[] for node in range(num_nodes)]
        self.in_degrees = Counter()
        self.num_nodes = num_nodes

    def add_edge(self, u: int, v: int):
        self.adj[u].append(v)
        self.in_degrees[v] += 1


class Solution:
    def sortItems(self, n: int, m: int, group: List[int], before_items: List[List[int]]) -> List[int]:
        for item in range(n):
            if group[item] == -1:
                group[item] = m
                m += 1

        groups_graph = Graph(m)
        items_graph = Graph(n)

        for item in range(n):
            item_group = group[item]

            for before_item in before_items[item]:
                before_item_group = group[before_item]

                if item_group == before_item_group:
                    items_graph.add_edge(before_item, item)
                else:
                    groups_graph.add_edge(before_item_group, item_group)

        try:
            groups_order = self.topological_sort(groups_graph)
            items_order = self.topological_sort(items_graph)
        except AssertionError:
            return []

        items_grouped = [[] for group in range(m)]
        for item in items_order:
            items_grouped[group[item]].append(item)

        result = []
        for group in groups_order:
            result.extend(items_grouped[group])

        return result

    def topological_sort(self, graph: Graph) -> List[int]:
        ordering = []
        queue = deque([node for node in range(graph.num_nodes) if graph.in_degrees[node] == 0])

        while queue:
            node = queue.popleft()
            ordering.append(node)

            for neighbor in graph.adj[node]:
                graph.in_degrees[neighbor] -= 1
                if graph.in_degrees[neighbor] == 0:
                    queue.append(neighbor)

        assert len(ordering) == graph.num_nodes

        return ordering

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