<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 deque, defaultdict
from typing import List
class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        new_group_index = m
        items_by_group = defaultdict(list)
        item_incoming_edges = [0] * n
        group_incoming_edges = [0] * (n + m)
        item_adjacency_list = defaultdict(list)
        group_adjacency_list = defaultdict(list)
        for i in range(n):
            if group[i] == -1:
                group[i] = new_group_index
                new_group_index += 1
            items_by_group[group[i]].append(i)
        for i in range(n):
            for before_item_index in beforeItems[i]:
                if group[i] == group[before_item_index]:
                    item_incoming_edges[i] += 1
                    item_adjacency_list[before_item_index].append(i)
                else:
                    group_incoming_edges[group[i]] += 1
                    group_adjacency_list[group[before_item_index]].append(group[i])
        all_groups = list(range(n + m))
        sorted_groups = self.topologicalSort(group_incoming_edges, group_adjacency_list, all_groups)
        if not sorted_groups:
            return []
        sorted_items = []
        for gid in sorted_groups:
            group_items = items_by_group[gid]
            sorted_group_items = self.topologicalSort(item_incoming_edges, item_adjacency_list, group_items)
            if len(sorted_group_items) != len(group_items):
                return []
            sorted_items.extend(sorted_group_items)
        return sorted_items
    def topologicalSort(self, incoming_edges: List[int], adjacency_list: defaultdict, items: List[int]) -> List[int]:
        queue = deque()
        for i in items:
            if incoming_edges[i] == 0:
                queue.append(i)
        sorted_items = []
        while queue:
            current = queue.popleft()
            sorted_items.append(current)
            for next_item in adjacency_list[current]:
                incoming_edges[next_item] -= 1
                if incoming_edges[next_item] == 0:
                    queue.append(next_item)
        return sorted_items if len(sorted_items) == len(items) else []

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