<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]:
        group_graph = defaultdict(set)
        grouped_items = defaultdict(set)
        in_degree_item = defaultdict(int)
        in_degree_group = defaultdict(int)

        for i in range(n):
            if group[i] == -1:
                group[i] = m
                m += 1
            grouped_items[group[i]].add(i)
        item_graph = [defaultdict(set) for _ in range(m)]

        for item in range(len(beforeItems)):
            nbrs = beforeItems[item]
            for nbr in nbrs:
                if group[nbr] == group[item]: 
                    in_degree_item[item] += 1
                    item_graph[group[item]][nbr].add(item)
                elif group[item] not in group_graph[group[nbr]]:
                    group_graph[group[nbr]].add(group[item])
                    in_degree_group[group[item]] += 1

        # print(group)
        # print(grouped_items)
        # print(group_graph)
        # print(in_degree_item)

        group_order = []
        que = deque([x for x in range(m) if in_degree_group[x] == 0])
        while que:
            node = que.popleft()
            group_order.append(node)
            for nbr in group_graph[node]:
                in_degree_group[nbr] -= 1
                if in_degree_group[nbr] == 0:
                    que.append(nbr)
        # print(group_order)
        if len(group_order) != m:
            return []
        final_order = []
        for g in group_order:
            que = deque([x for x in grouped_items[g] if in_degree_item[x] == 0])
            item_order = []
            while que:
                node = que.popleft()
                item_order.append(node)
                for nbr in item_graph[g][node]:
                    in_degree_item[nbr] -= 1
                    if in_degree_item[nbr] == 0:
                        que.append(nbr)
            # print(g, item_order)
            if len(item_order) != len(grouped_items[g]):
                return []
            final_order.extend(item_order)
        return final_order

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