<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_id = m

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

        item_graph = [[] for _ in range(n)]
        item_in_degree = [0] * n

        group_graph = [[] for _ in range(group_id)]
        group_in_degree = [0] * group_id

        # build edges
        for node in range(n):
            for parent in beforeItems[node]:
                item_graph[parent].append(node)
                item_in_degree[node] += 1

                # diff groups?
                if group[parent] != group[node]:
                    group_graph[group[parent]].append(group[node])
                    group_in_degree[group[node]] += 1

        # run topo sort on items(without group ordering) and groups
        def topo_sort(graph, in_degree):
            queue = deque([node for node in range(len(graph)) if in_degree[node] == 0])
            sorted_list = []

            while queue:
                node = queue.popleft()
                sorted_list.append(node)
                for neighbor in graph[node]:
                    in_degree[neighbor] -= 1
                    if in_degree[neighbor] == 0:
                        queue.append(neighbor)

            # invalid sorting
            if len(sorted_list) != len(graph):
                return []
            return sorted_list
        
        item_order = topo_sort(item_graph, item_in_degree)
        group_order = topo_sort(group_graph, group_in_degree)

        if not item_order or not group_order:
            return []
        ordered_groups = defaultdict(list)
        # make a hash map of group -> items
        for item in item_order:
            ordered_groups[group[item]].append(item)
        
        ans = []
        for group_idx in group_order:
            ans.extend(ordered_groups[group_idx])
        return ans

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