<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]:
        before = {i:set() for i in range(n)}
        after = {i:set() for i in range(n)}
        beforeG = {i:set() for i in range(m)}
        afterG = {i:set() for i in range(m)}
        groups = {i:set() for i in range(m)}
        qg = collections.deque()
        lazy = collections.deque()
        for i in range(n):
            if group[i] != -1:
                groups[group[i]].add(i)
            for x in beforeItems[i]:
                before[i].add(x)
                after[x].add(i)
                if group[x] != group[i] and group[x] != -1 and group[i] != -1:
                    beforeG[group[i]].add(group[x])
                    afterG[group[x]].add(group[i])
        for i in range(n):
            if group[i] == -1 and not before[i]:
                lazy.append(i)

        for i in range(m):
            if not beforeG[i]:
                qg.append(i)
        ans = []
        while qg:
            while lazy:
                i = lazy.popleft()
                ans.append(i)
                for j in after[i]:
                    before[j].remove(i)
                    if not before[j] and group[j] == -1:
                        lazy.append(j)
            g = qg.popleft()
            q = collections.deque()
            for member in groups[g]:
                if not before[member]:
                    q.append(member)
            while q:
                i = q.popleft()
                ans.append(i)
                groups[g].remove(i)
                for j in after[i]:
                    before[j].remove(i)
                    if not before[j]:
                        if group[j] == g:
                            q.append(j)
                        if group[j] == -1:
                            lazy.append(j)     
            if groups[g]:
                return []
            for p in afterG[g]:
                beforeG[p].remove(g)
                if not beforeG[p]:
                    qg.append(p)
        while lazy:
            i = lazy.popleft()
            ans.append(i)
            for j in after[i]:
                before[j].remove(i)
                if not before[j] and group[j] == -1:
                    lazy.append(j)
        return ans if len(ans) == n else []
420