<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]:
        for i in range(n):
            if group[i] == -1:
                group[i] = m
                m += 1

        itemGraph, itemDegree = defaultdict(list), defaultdict(int)
        groupGraph, groupDegree = defaultdict(list), defaultdict(int)
        groupToItems = defaultdict(list)

        for crs, gCrs in enumerate(group):
            groupToItems[gCrs].append(crs)
            for pre in beforeItems[crs]:
                gPre = group[pre]
                if gCrs == gPre:
                    itemGraph[pre].append(crs)
                    itemDegree[crs] += 1
                else:
                    groupGraph[gPre].append(gCrs)
                    groupDegree[gCrs] += 1

        def topSort(items, graph, degree):
            q = deque()
            for i in items:
                if degree[i] == 0:
                    q.append(i)
            sortedList = []
            while q:
                i = q.popleft()
                sortedList.append(i)
                for nei in graph[i]:
                    degree[nei] -= 1
                    if degree[nei] == 0:
                        q.append(nei)
            return sortedList if len(sortedList) == len(items) else []

        
        sortedGroups = topSort(list(range(m)), groupGraph, groupDegree)
        if not sortedGroups:
            return []

        res = []
        for grp in sortedGroups:
            items = groupToItems[grp]
            sortedItems = topSort(items, itemGraph, itemDegree)
            if len(items) != len(sortedItems):
                return []
            res.extend(sortedItems)

        return res
        
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