<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]:
        from collections import defaultdict, deque

        def topological_sort(nodes, pres, nxts):
            in_degrees = {node: 0 for node in nodes}
            for node in nodes:
                for next_node in nxts[node]:
                    in_degrees[next_node] += 1
            q = deque([node for node in nodes if in_degrees[node] == 0])
            res = []
            while q:
                node = q.popleft()
                res.append(node)
                for next_node in nxts[node]:
                    in_degrees[next_node] -= 1
                    if in_degrees[next_node] == 0:
                        q.append(next_node)
            if len(res) == len(nodes):
                return res
            else:
                return []

        # Step 1: Assign unique group IDs to ungrouped items
        ungrouped_id = m
        for i in range(n):
            if group[i] == -1:
                group[i] = ungrouped_id
                ungrouped_id += 1

        # Step 2: Create group and item dependencies
        group_items = defaultdict(list)
        group_pres = defaultdict(set)
        group_nxts = defaultdict(set)
        item_pres = defaultdict(set)
        item_nxts = defaultdict(set)

        for i in range(n):
            group_items[group[i]].append(i)
            for j in beforeItems[i]:
                if group[j] != group[i]:
                    group_nxts[group[j]].add(group[i])
                    group_pres[group[i]].add(group[j])
                else:
                    item_nxts[j].add(i)
                    item_pres[i].add(j)

        # Step 3: Topologically sort items within each group
        sorted_items_within_group = {}
        for g in group_items:
            sorted_items = topological_sort(group_items[g], item_pres, item_nxts)
            if not sorted_items:
                return []  # Cycle detected within a group
            sorted_items_within_group[g] = sorted_items

        # Step 4: Topologically sort the groups
        all_groups = list(group_items.keys())
        sorted_groups = topological_sort(all_groups, group_pres, group_nxts)
        if not sorted_groups:
            return []  # Cycle detected between groups

        # Step 5: Combine the sorted items
        result = []
        for g in sorted_groups:
            result.extend(sorted_items_within_group[g])

        return result
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