<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]:
        grouping = [[] for _ in range(m)]
        for ind, g in enumerate(group):
            if g != -1:
                grouping[g].append(ind)

        adj_list = [[] for _ in range(n)]
        in_degrees = [0] * n # indegrees of a node
        group_indegrees = [0] * m # external indegrees of group

        for node, before_list in enumerate(beforeItems):
            for before in before_list:
                adj_list[before].append(node)
                in_degrees[node] += 1
                if group[node] != -1 and group[node] != group[before]: # external indegree
                    group_indegrees[group[node]] += 1

        sol = []

        queue = deque()

        for node, degrees in enumerate(in_degrees):
            if not degrees and group[node] == -1:
                queue.append([node])
        
        for g, degrees in enumerate(group_indegrees):
            if not degrees:
                queue.append(grouping[g])
        
        while queue:
            g = queue.popleft()
            group_sorted = []

            kahn_queue = deque()
            for node in g:
                if not in_degrees[node]:
                    kahn_queue.append(node)
            
            while kahn_queue:
                node = kahn_queue.popleft()
                group_sorted.append(node)
                for neighbor in adj_list[node]:
                    in_degrees[neighbor] -= 1
                    neighbor_group = group[neighbor]
                    if neighbor_group == -1: # neighbor has no group
                        if not in_degrees[neighbor]:
                            queue.append([neighbor])
                    elif group[node] != neighbor_group: # different group
                        group_indegrees[neighbor_group] -= 1
                        if not group_indegrees[neighbor_group]: # group indegrees 0 
                            queue.append(grouping[neighbor_group])
                    else: # same group 
                        if not in_degrees[neighbor]:
                            kahn_queue.append(neighbor)
            
            if len(group_sorted) != len(g):
                return []
            
            sol += group_sorted

        return sol if len(sol) == n else []
        
305