<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 topologicalSort(self, num_nodes, sucessors, predecessor_count):
        q = collections.deque()
        order = []

        for node in range(num_nodes):
            if predecessor_count[node] == 0:
                q.append(node)
        
        while q:
            node = q.popleft()
            order.append(node)
            for successor in sucessors[node]:
                predecessor_count[successor] -= 1
                if predecessor_count[successor] == 0:
                    q.append(successor)
        
        if len(order) == num_nodes:
            return order
        return []


    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
        
        successor_items, successor_groups = [[] for _ in range(n)], [[] for _ in range(m)]
        predecessor_items, predecessor_groups = [0] * n, [0] * m

        for item in range(n):
            currentGroup = group[item]
            for beforeItem in beforeItems[item]:
                beforeGroup = group[beforeItem]
                if currentGroup == beforeGroup:
                    successor_items[beforeItem].append(item)
                    predecessor_items[item] += 1
                else:
                    successor_groups[beforeGroup].append(currentGroup)
                    predecessor_groups[currentGroup] += 1

        order_items, order_groups = self.topologicalSort(n, successor_items, predecessor_items), self.topologicalSort(m, successor_groups, predecessor_groups)

        if not order_items or not order_groups:
            return []
        
        item_grouped = [[] for _ in range(m)]
        for item in order_items:
            item_grouped[group[item]].append(item)
        
        result = []

        for group in order_groups:
            result.extend(item_grouped[group])
        
        return result
33706