<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]:
        def topologicalSort(degrees, graph, items):
            queue = deque([item for item in items if degrees[item] == 0])
            res = []
            while queue:
                curr = queue.popleft()
                res.append(curr)
                for nei in graph[curr]:
                    degrees[nei] -= 1
                    if degrees[nei] == 0:
                        queue.append(nei)
            return res if len(res) == len(items) else []


        # Assign each standalone item a unique group member
        index = m
        group_to_items = [[] for _ in range(n + m)]
        for item, grp in enumerate(group):
            if grp == -1:
                group[item] = index
                index += 1
            group_to_items[group[item]].append(item)
        
        # Initialize degree and graph for items and groups:
        item_degree = [0] * n
        group_degree = [0] * (n + m)
        item_graph = [[] for _ in range(n)]
        group_graph = [[] for _ in range(n + m)]

        # Build Graph for items and group
        for i, grp in enumerate(group):
            for j in beforeItems[i]:
                gj = group[j]
                if grp == gj: # Same group
                    item_degree[i] += 1
                    item_graph[j].append(i)
                else:
                    if grp not in group_graph[j]:
                        group_degree[grp] += 1
                        group_graph[gj].append(grp)

        # Perform topological sort
        # all_items = [] * (n + m)
        # for i in range(n + m):
        #     all_items.append(i)

        group_order = topologicalSort(group_degree, group_graph, list(range(n + m)))
        if not group_order:
            return []

        final_order = []
        # For each group in the sorted order perform topological sort in the items
        for group_index in group_order:
            items_in_group = group_to_items[group_index]
            item_order = topologicalSort(item_degree, item_graph, items_in_group)
            if len(items_in_group) != len(item_order):
                return []
            final_order.extend(item_order)
        return final_order


        

        
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