<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]:
        #define toposort function
        
        def topoSort(graphDct, degreeDct):
            #graphDct = {0:[], 1:[6], 2:[5], 3:[6], 4:[3,6], 5:[], 6:[], 7:[]}
            #number of outgoing edges
            #degreeDct =  {0:0, 1:0, 2:0, 3:1, 4:0, 5:1, 6:3, 7:0}
            stk = [] 
            visit = [] #[7]
            for node, degree in degreeDct.items():
                if degree == 0:
                    stk.append(node)
            while stk: #[0,5,6]
                currNode = stk.pop()
                visit.append(currNode)
                for neighbor in graphDct[currNode]:
                    degreeDct[neighbor] -= 1
                    if degreeDct[neighbor] == 0:
                        stk.append(neighbor)
            return [] if len(visit) != len(degreeDct) else visit
    
        #change -1 group numbers
        groupNum = m
        for i in range(len(group)):
            if group[i] == -1:
                group[i] = groupNum
                groupNum += 1
        # print(f'group:{group}')
        itemGraph = {i:[] for i in range(n)} 
        #in solution, itemGraph = {prereq:[courses to take after]}, itemDegree = {course:number incoming edges}
        groupGraph = {i:[] for i in range(groupNum)}
        itemDegree = {i:0 for i in range(n)}
        groupDegree = {i:0 for i in range(groupNum)}
        for course, prereqs in enumerate(beforeItems):
            for prereq in prereqs:
                itemGraph[prereq].append(course)
                itemDegree[course] += 1
                if group[prereq] != group[course]:
                    groupGraph[group[prereq]].append(group[course])
                    groupDegree[group[course]] += 1

        # print(f'itemGraph:{itemGraph}')
        # print(f'groupGraph:{groupGraph}')
        # print(f'itemDegree:{itemDegree}')
        # print(f'groupDegree:{groupDegree}')


        itemLst = topoSort(itemGraph, itemDegree)
        grouplst = topoSort(groupGraph, groupDegree)
        # print(f'itemLst{itemLst}')
        # print(f'grouplst{grouplst}')
        groups = {g:[] for g in range(groupNum)}
        # print(f'before groups:{groups}')
        #for each item in item toposort :
        for item in itemLst:
            #add to defaultdict for groups
            groups[group[item]].append(item)
        # print(f'after groups:{groups}')
        #concatenate together in order of group topo sort
        res = []
        for group in grouplst:
            res.extend(groups[group])
        return res
        
    
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