<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>

"""
___
Goal:
___

Return any solution if there exists one, otherwise return empty list if no solution

___
Examples:
___

Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3,6],[],[],[]]
Output: [6,3,4,1,5,2,0,7]



Input: n = 8, m = 2, group = [-1,-1,1,0,0,1,0,-1], beforeItems = [[],[6],[5],[6],[3],[],[4],[]]
Output: []

___
Assumptions:
___

1. we are returning sorted list of items

2. items that belong in same group are next to eachother in sorted list

3. beforeItems[i] is list of items are dependencies of i-th item of sorted list
    - i.e. 


4. 0 <= i < n items

5. -1 <= group[i] <= m - 1 

___
PsuedoCode: T: O(vlog(v) + e), S: O(v)
___

Topological Sort:
    - This determines if there is a possible solution by detecting cycle
    - topology has consequence on ordering:
        - imposes dependency structure at two levels of abstractions:
            1. intergroups: some items have precedence over other items
            2. intragroups: some groups have precedence over other groups


1. Use Topological sort to get group orderings
    - intragroup

2. for each group in orderings:
    - do topological sortings and add items
    - finds intergroup


"""


class Solution:
    def sortItems(
        self, n: int, m: int, group: List[int], beforeItems: List[List[int]]
    ) -> List[int]:

        def topologicalSort(nodes, pred, subsequent):
            order = []
            queue = deque(node for node in nodes if not pred[node])

            while queue:
                node = queue.popleft()
                order.append(node)

                for s in subsequent[node]:
                    pred[s].discard(node)
                    if not pred[s]:
                        queue.append(s)
            return order if len(order) == len(nodes) else []

        group_to_items = {}
        new_group = m

        for item in range(n):
            if reassign_group := group[item] == -1:
                group[item] = new_group
                new_group += 1

            if group[item] not in group_to_items:
                group_to_items[group[item]] = set()

            group_to_items[group[item]].add(item)

        intra_indegree = defaultdict(set)
        intra_subsequent = defaultdict(set)
        inter_indegree = defaultdict(set)
        inter_subsequent = defaultdict(set)

        for item in range(n):
            for before in beforeItems[item]:
                if same_item_membership := group[item] == group[before]:
                    intra_indegree[item].add(before)
                    intra_subsequent[before].add(item)
                else:
                    inter_indegree[group[item]].add(group[before])
                    inter_subsequent[group[before]].add(group[item])

        groups_order = topologicalSort(
            list(group_to_items.keys()), inter_indegree, inter_subsequent
        )
        if not groups_order:
            return []

        items_order = []
        for grp in groups_order:
            order = topologicalSort(group_to_items[grp], intra_indegree, intra_subsequent)
            if not order:
                return []
            items_order.extend(order)

        return items_order

343