Sort Items by Groups Respecting Dependencies
<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://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1200-1299/1203.Sort%20Items%20by%20Groups%20Respecting%20Dependencies/images/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>
Depth-First Search; Breadth-First Search; Graph; Topological Sort
class Solution:
    def sortItems(
        self, n: int, m: int, group: List[int], beforeItems: List[List[int]]
    ) -> List[int]:
        def topo_sort(degree, graph, items):
            q = deque(i for _, i in enumerate(items) if degree[i] == 0)
            res = []
            while q:
                i = q.popleft()
                res.append(i)
                for j in graph[i]:
                    degree[j] -= 1
                    if degree[j] == 0:
                        q.append(j)
            return res if len(res) == len(items) else []

        idx = m
        group_items = [[] for _ in range(n + m)]
        for i, g in enumerate(group):
            if g == -1:
                group[i] = idx
                idx += 1
            group_items[group[i]].append(i)

        item_degree = [0] * n
        group_degree = [0] * (n + m)
        item_graph = [[] for _ in range(n)]
        group_graph = [[] for _ in range(n + m)]
        for i, gi in enumerate(group):
            for j in beforeItems[i]:
                gj = group[j]
                if gi == gj:
                    item_degree[i] += 1
                    item_graph[j].append(i)
                else:
                    group_degree[gi] += 1
                    group_graph[gj].append(gi)

        group_order = topo_sort(group_degree, group_graph, range(n + m))
        if not group_order:
            return []
        ans = []
        for gi in group_order:
            items = group_items[gi]
            item_order = topo_sort(item_degree, item_graph, items)
            if len(items) != len(item_order):
                return []
            ans.extend(item_order)
        return ans
