<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]:
        for i in range(n):
            if group[i] == -1:
                group[i] = m
                m += 1

        # Stage 1: sort in each group
        group_to_items = [[] for i in range(m)]
        for i in range(n):
            group_to_items[group[i]].append(i)
        
        for group_i in range(m):
            if len(group_to_items[group_i]) <= 1:
                continue
            item_set = set(group_to_items[group_i])
            group_adj = {i: [j for j in beforeItems[i] if j in item_set] for i in item_set}
            in_degrees = {i: 0 for i in item_set}
            for i in group_adj:
                for j in group_adj[i]:
                    in_degrees[j] += 1
            visited = []
            to_visit = [i for i in item_set if in_degrees[i] == 0]
            while len(to_visit) > 0:
                current = to_visit.pop()
                visited.append(current)
                for next_node in group_adj[current]:
                    in_degrees[next_node] -= 1
                    if in_degrees[next_node] == 0:
                        to_visit.append(next_node)
            if len(visited) < len(item_set):
                return []
            group_to_items[group_i] = list(reversed(visited))
        
        #print(group_to_items)

        # Stage 2: sort groups
        item_to_group = {}
        for group_i in range(m):
            for j in group_to_items[group_i]:
                item_to_group[j] = group_i
        adj = [set() for i in range(m)]
        for i in range(n):
            for j in beforeItems[i]:
                if item_to_group[j] != item_to_group[i]:
                    adj[item_to_group[j]].add(item_to_group[i])

        #print(adj)

        in_degrees = [0 for i in range(m)]
        for i in range(m):
            for j in adj[i]:
                in_degrees[j] += 1
        visited = []
        to_visit = [i for i in range(m) if in_degrees[i] == 0]
        while len(to_visit) > 0:
            current = to_visit.pop()
            visited.append(current)
            for i in adj[current]:
                in_degrees[i] -= 1
                if in_degrees[i] == 0:
                    to_visit.append(i)

        #print(visited)

        if len(visited) < m:
            return []
        result = []
        for i in visited:
            result.extend(group_to_items[i])
        return result
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