<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 top_sort(nodes, edges):
    indeg = defaultdict(int)
    output = []
    nodes = set(nodes)
    for x in nodes:
        for y in edges[x]:
            if y in nodes: indeg[y] += 1
    q = deque([node for node in nodes if not indeg[node]])
    while q:
        curr = q.popleft()
        output += [curr]
        for nxt in edges[curr]:
            indeg[nxt] -= 1
            if not indeg[nxt]: q += [nxt]
    return output

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        edges = defaultdict(list)
        groups = [[] for _ in range(m+1)]
        for i, before in enumerate(beforeItems):
            for b in before:
                edges[b] += [i]
                I, B = group[i], group[b]
                if I != -1 and B != -1:
                    if I != B: edges[B+n] += [I+n]
                elif I != -1: edges[b] += [I+n]
                elif B != -1: edges[B+n] += [i]
        naive = top_sort(range(n), edges)
        if len(naive) != n: return []
        for node in naive: groups[group[node]] += [node]
        strict = top_sort(list(range(n, n+m))+groups[-1], edges)
        output = []
        for s in strict:
            if s >= n: output += groups[s-n]
            else: output += [s]
        if len(output) != n: return []
        return output
                    
        
41806