<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]:
        graph = {i:[] for i in range(n)}
        inDegreesItems = {i:0 for i in range(n)}
        groupGraph = {j:set() for j in range(m)}
        inDegreesGroups = {j:0 for j in range(m)}
        numGroups = m
        
        for i in range(n):
            if group[i] == -1: 
                group[i] = numGroups
                numGroups += 1
                groupGraph[group[i]] = set()
                inDegreesGroups[group[i]] = 0

            for j in beforeItems[i]:
                if group[j] == -1: 
                    group[j] = numGroups
                    numGroups += 1
                    groupGraph[group[j]] = set()
                    inDegreesGroups[group[j]] = 0
                graph[j].append(i)
                inDegreesItems[i] += 1
                if group[i] != group[j] and group[i] not in groupGraph[group[j]]:
                    groupGraph[group[j]].add(group[i])
                    inDegreesGroups[group[i]] += 1

        def topoSort(graph,inDegrees,num):
            itemQ = []
            for i in range(num):
                if inDegrees[i] == 0:
                    heapq.heappush(itemQ, i)
            
            itemOrder = []
            visitedItems = set()
            while itemQ:
                for _ in range(len(itemQ)):
                    curElement = heapq.heappop(itemQ)
                    itemOrder.append(curElement)
                    visitedItems.add(curElement)
                    for nei in graph[curElement]:
                        inDegrees[nei] -= 1
                        if inDegrees[nei] == 0:
                            heapq.heappush(itemQ, nei)
            if len(visitedItems) != num:
                return []
            return itemOrder

        itemOrder = topoSort(graph,inDegreesItems,n)
        groupOrder = topoSort(groupGraph,inDegreesGroups,numGroups)
        if not itemOrder or not groupOrder:
            return []

        resultGroups = {i:[] for i in groupOrder}
        for item in itemOrder:
            resultGroups[group[item]].append(item)
        result = []
        for group in groupOrder:
            for item in resultGroups[group]:
                result.append(item)
        return result
48893