<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]:
        # build map of group to all items in given group
        group_members = defaultdict(list)
        for i, g in enumerate(group):
            # if item belongs to no group, we put it into a group of its own; this
            # new group will have ID `m+i` to ensure the item is the sole member
            if g == -1: g = group[i] = i + m
            group_members[g].append(i)

        # use the created `group_members` & `beforeItems` to build a directed graph of group relations
        group_conditions = []
        for g, items in group_members.items():
            for i in items:
                for pi in beforeItems[i]:
                    pg = group[pi]
                    if pg == g: continue    # items belonging to the same group will not contribute
                    group_conditions.append([pg, g])

        # topologically sort the groups based on `group_conditions` built above
        group_sorted, group_order = Solution.topological_sort(group_members.keys(), group_conditions)
        if not group_sorted: return []  # group cannot be sorted due to cycle, thus items are unsortable
        group_order_map = {g: o for (o, g) in enumerate(group_order)}

        # build a directed graph of item relations
        item_conditions = []
        for it, prevs in enumerate(beforeItems):
            for pi in prevs: item_conditions.append((pi, it))

        # topologically sort the items based on the `item_conditions` built above
        item_sorted, items = Solution.topological_sort(range(n), item_conditions)
        if not item_sorted: return []

        # while maintaining individual item orders, sort all elements by `group_order_map`
        # thus placing all items in the same group adjacent to each other as a single chunk
        items.sort(key=lambda i: group_order_map[group[i]])
        return items


    @staticmethod
    def topological_sort(elements, conditions):
        edges, in_deg_cnts = defaultdict(list), Counter()
        for a, b in conditions:
            edges[a].append(b)
            in_deg_cnts[b] += 1
        
        S, L = collections.deque(i for i in elements if in_deg_cnts[i] == 0), []
        while S:
            a = S.popleft()
            L.append(a)
            for b in edges[a]:
                in_deg_cnts[b] -= 1
                if in_deg_cnts[b] == 0: S.append(b)
        
        return (False, []) if any(in_deg_cnts.values()) else (True, L)
382