<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]:
        def get_top_order(adj, indegrees):
            top_order = []
            stack = [i for i in range(len(indegrees)) if indegrees[i] == 0]
            print(indegrees)
            while stack:
                top_order.append(parent := stack.pop())

                for child in adj[parent]:
                    indegrees[child] -= 1

                    if indegrees[child] == 0:
                        stack.append(child)

            return top_order if len(top_order) == len(adj) else []

        for u in range(len(group)):
            if group[u] == -1:
                group[u] = m
                m += 1

        adj_items, indegree_items = defaultdict(list), Counter({i: 0 for i in range(n)})
        adj_groups, indegree_groups = defaultdict(list), Counter({i: 0 for i in range(m)})
        for u in range(n):
            for v in beforeItems[u]:
                adj_items[v].append(u)
                indegree_items[u] += 1

                if group[u] != group[v]:
                    adj_groups[group[v]].append(group[u])
                    indegree_groups[group[u]] += 1

        item_order = get_top_order(adj_items, indegree_items)
        group_order = get_top_order(adj_groups, indegree_groups)

        if not item_order or not group_order:
            return []

        order_within_group = defaultdict(list)
        for v in item_order:
            order_within_group[group[v]].append(v)

        res = []
        for g in group_order:
            res += order_within_group[g]

        return res

369