<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>

def kahn_sort(vertices, edges):
	successors = {v:[] for v in vertices}
	predecessor_counts = Counter()
	for a,b in edges:
		successors[a].append(b)	
		predecessor_counts[b] += 1

	order = []
	stack = [a for a in successors if predecessor_counts[a] == 0]
	while stack:
		a = stack.pop()
		order.append(a)
		for b in successors[a]:
			predecessor_counts[b] -= 1
			if predecessor_counts[b] == 0:
				stack.append(b)

	if any(predecessor_counts.values()):
		return []
	return order

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        groups = defaultdict(list)
        for i,g in enumerate(group):
            if g == -1:
                groups[m+i].append(i)
            else:
                groups[g].append(i)
        group_edges = defaultdict(list)
        edges = []
        for i,before in enumerate(beforeItems):
            for j in before:
                if group[i] == group[j] and group[i] != -1:
                    group_edges[group[i]].append((j,i))
                else:
                    edges.append((m+j if group[j] == -1 else group[j],m+i if group[i] == -1 else group[i]))
        for g,l in groups.items():
            if group_edges[g]:
                if not (order := kahn_sort(l,group_edges[g])):
                    return []
                groups[g] = order
        return list(chain.from_iterable(groups[g] for g in kahn_sort(groups.keys(),edges)))
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