<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]:
        
        groupID = m
        for i in range(n):
            if group[i] == -1:
                group[i] = groupID
                groupID += 1
        # groupID will be the max num of groups
        group_indegree = defaultdict(int)       # number of pres of the group
        group_nei = defaultdict(list)           # neighbor of the group   a -- b: need to work on a before b

        project_indegree = defaultdict(int)     # number of pres of the project
        project_nei = defaultdict(list)         # neighbor of a project   a -- b: need to work on a before b

        group_project = defaultdict(list)       #  groupID -- projects in the group

        for p in range(n):
            group_project[group[p]].append(p)
            for pre in beforeItems[p]:
                if group[pre] != group[p]:   # pre and current project are from different groups, need to work on group topology
                    group_indegree[group[p]] += 1
                    group_nei[group[pre]].append(group[p])
                else:           # pre and project are in same group, work on the project topology
                    project_indegree[p] += 1
                    project_nei[pre].append(p)

        res = []
        # toposort the groups
        group_queue = self.toposort([i for i in range(groupID)], group_indegree, group_nei)
        print(group_queue)
        if len(group_queue) != groupID:
            return []

        for group in group_queue:
            project_queue = self.toposort(group_project[group], project_indegree, project_nei)
            if len(project_queue) != len(group_project[group]):
                return []
            
            res += project_queue

        return res


    def toposort(self, items, indegree, neighbor):

        q = deque()
        res = []

        for item in items:
            if item not in indegree:
                q.append(item)
        
        while q:
            node = q.popleft()
            res.append(node)
            for nei in neighbor[node]:
                indegree[nei] -= 1
                if indegree[nei] == 0:
                    q.append(nei)
        
        return res
37081