<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, deque
from typing import List

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        # Adjust group numbers for items that belong to no group
        for i in range(n):
            if group[i] == -1:
                group[i] = m
                m += 1

        # Create adjacency lists and in-degree counts for items and groups
        item_graph = defaultdict(list)
        group_graph = defaultdict(list)
        item_indegree = [0] * n
        group_indegree = [0] * m

        for i in range(n):
            for before in beforeItems[i]:
                item_graph[before].append(i)
                item_indegree[i] += 1
                if group[i] != group[before]:
                    group_graph[group[before]].append(group[i])
                    group_indegree[group[i]] += 1

        # Kahn's algorithm for topological sorting
        def topological_sort(graph, indegree, size):
            zero_indegree = deque([i for i in range(size) if indegree[i] == 0])
            sorted_order = []
            while zero_indegree:
                node = zero_indegree.popleft()
                sorted_order.append(node)
                for neighbor in graph[node]:
                    indegree[neighbor] -= 1
                    if indegree[neighbor] == 0:
                        zero_indegree.append(neighbor)
            if len(sorted_order) == size:
                return sorted_order
            else:
                return []

        # Perform topological sort on groups and items
        group_order = topological_sort(group_graph, group_indegree, m)
        if not group_order:
            return []

        item_order = topological_sort(item_graph, item_indegree, n)
        if not item_order:
            return []

        # Place items in the order respecting groups and items within groups
        group_to_items = defaultdict(list)
        for item in item_order:
            group_to_items[group[item]].append(item)

        result = []
        for g in group_order:
            result.extend(group_to_items[g])

        return result

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