<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 topologicalSort(self, elements, elementDependents, elementInDegree):
        ordering = []
        q = deque(element for element in elements if elementInDegree[element] == 0)
        while len(q):
            element = q.popleft()
            for child in elementDependents[element]:
                elementInDegree[child] -= 1
                if elementInDegree[child] == 0:
                    q.append(child)
            ordering.append(element)
        return ordering if len(ordering) == len(elements) else []

    def sortItems(
        self, n: int, m: int, group: List[int], beforeItems: List[List[int]]
    ) -> List[int]:
        groupToItems = defaultdict(list)  # Build groupToItems Map
        inDegree = [0] * n  # Indegree of items
        itemDependents = defaultdict(list) # Item -> Items that depend on it
        groupInDegree = defaultdict(
            lambda: 0
        )  # Indegree of groups (how many dependencies do they have)
        groupDependents = defaultdict(list)  # Group -> Groups that depend on it
        newGroup = m
        # Build Maps and assign each item with no group a group of its own
        # to simplify processing
        for item in range(n):
            if group[item] == -1:
                group[item] = newGroup
                newGroup += 1
            groupToItems[group[item]].append(item)

            for beforeItem in beforeItems[
                item
            ]: 
                if group[beforeItem] == -1: # Assign new group to group-less items
                    group[beforeItem] = newGroup
                    newGroup += 1

                # If they belong to differnet groups,
                # They will be handled by groups topological sort
                # Update group dependencies 
                if group[beforeItem] != group[item]: 
                    groupDependents[group[beforeItem]].append(group[item])
                    groupInDegree[group[item]] += 1
                else:
                    # If they belong to the same group,
                    # they will be handled by same Group topological sort
                    # update the item dependencies
                    itemDependents[beforeItem].append(item)
                    inDegree[item] += 1

        ## Top Sort for groups
        groupOrdering = self.topologicalSort(
            range(newGroup), groupDependents, groupInDegree
        )
        if groupOrdering == []:
            return []

        res = []
        for g in groupOrdering:
            items = groupToItems[g]
            if len(items) == 0:
                continue
            output = self.topologicalSort(items, itemDependents, inDegree)
            res = res + output

        return res if len(res) == n else []

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