<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]:
        # first, topological sort on groups 0 - (m - 1)
        group_adj = {i : set() for i in range(m)}
        for g, before in zip(group, beforeItems):
            if g == -1:
                continue
            for b in before:
                if group[b] != g:
                    group_adj[g].add(group[b])
        
        group_visited = set()
        group_path = set()
        group_sort = []

        def dfs_group(i: int) -> bool:
            if i in group_path:
                return False
            if i in group_visited or i == -1:
                return True
            group_visited.add(i)
            group_path.add(i)
            for n in group_adj[i]:
                if not dfs_group(n):
                    return False
            group_path.remove(i)
            group_sort.append(i)
            return True
        
        for i in range(m):
            if not dfs_group(i):
                return []
        
        
        group_lists = {i : [] for i in range(-1, m)}
        for i, g in enumerate(group):
            group_lists[g].append(i)

        # now topological sort on item groups, in order of group lists
        # but first, add predecessors with group -1 to all items in a 
        # group if any item in the group has a predecessor -1
        beforeItems = [set(beforeItems[i]) for i in range(n)]
        for i, before in enumerate(beforeItems):
            for b in before:
                if group[b] == -1 and group[i] != -1:
                    for gm in group_lists[group[i]]:
                        beforeItems[gm].add(b)
        
        item_path = set()
        item_visited = set()
        item_arr = []

        def dfs_item(i: int) -> bool:
            if i in item_path:
                return False
            if i in item_visited:
                return True
            item_path.add(i)
            item_visited.add(i)
            for n in beforeItems[i]:
                if not dfs_item(n):
                    return False
            item_arr.append(i)
            item_path.remove(i)
            return True

        for g in group_sort:
            for j in group_lists[g]:
                if not dfs_item(j):
                    return []
        for j in group_lists[-1]:
            if not dfs_item(j):
                return []
        return item_arr
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