<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>

from collections import deque, defaultdict

class Solution:
    def topo_sort(self, adj, deg, sz):
        # Kahn's algorithm
        q = deque()
        for i in range(sz):
            if deg[i] == 0:
                q.append(i)
        res = []
        while q:
            x = q.popleft()
            res.append(x)
            for y in adj[x]:
                deg[y] -= 1
                if deg[y] == 0:
                    q.append(y)
        
        for i in range(sz):
            if deg[i] > 0:
                return []
        return res

    def sortItems(self, n, m, group, beforeItems):
        # Isolated items are treated as their own groups
        for i in range(n):
            if group[i] == -1:
                group[i] = m
                m += 1

        adj_item = defaultdict(list)
        deg_group = [0] * m
        deg_item = [0] * n
        adj_group_set = [set() for _ in range(m)]  # Avoid multiple edges

        for i in range(n):
            to_i = group[i]
            for from_item in beforeItems[i]:
                from_j = group[from_item]
                if to_i != from_j and to_i not in adj_group_set[from_j]:
                    adj_group_set[from_j].add(to_i)
                    deg_group[to_i] += 1
                adj_item[from_item].append(i)
                deg_item[i] += 1

        sort_group = self.topo_sort(adj_group_set, deg_group, m)
        sort_item = self.topo_sort(adj_item, deg_item, n)

        if not sort_group or not sort_item:
            return []

        item_gp = [[] for _ in range(m)]
        for i in sort_item:
            item_gp[group[i]].append(i)

        ans = []
        for i in sort_group:
            for item in item_gp[i]:
                ans.append(item)

        return ans

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