<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 topoSort(nodes, graph):
            res = []
            visited, path = set(), set()

            def dfs(node):
                if node in path:
                    return False
                if node in visited:
                    return True

                path.add(node)
                for nei in graph[node]:
                    if not dfs(nei):
                        return False
                path.remove(node)
                visited.add(node)
                res.append(node)
                return True

            for node in nodes:
                dfs(node)

            return res[::-1]

        groups = collections.defaultdict(list)
        groupId = m
        for i in range(n):
            if group[i] == -1:
                group[i] = groupId
                groupId += 1
            groups[group[i]].append(i)

        itemGraph = collections.defaultdict(list)
        for v, uList in enumerate(beforeItems):
            for u in uList:
                if group[v] == group[u]:
                    itemGraph[u].append(v)

        sortedGroups = collections.defaultdict(list)
        for groupId in groups:
            sortedGroup = topoSort(groups[groupId], itemGraph)
            if len(sortedGroup) != len(groups[groupId]):
                return []
            sortedGroups[groupId] = sortedGroup

        groupGraph = collections.defaultdict(list)
        for v, uList in enumerate(beforeItems):
            for u in uList:
                if group[v] != group[u]:
                    groupGraph[group[u]].append(group[v])

        groupOrder = topoSort(set(group), groupGraph)
        if len(groupOrder) != len(set(group)):
            return []

        res = []
        for groupId in groupOrder:
            res.extend(sortedGroups[groupId])
        return res
395