<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 defaultdict

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        for i in range(len(group)):
            if group[i] == -1: 
                group[i] = m
                m += 1 

        graph_groups = defaultdict(set)
        for i in range(m):
            graph_groups[i] = []
        
        graph_items = defaultdict(set)
        for i in range(n):
            graph_items[i] = []

        graph_groups_indegree = self.make_graph_groups(n, group, beforeItems, graph_groups)
        graph_items_indegree = self.make_graph_items(n, group, beforeItems, graph_items)

        group_list = self.topo_sort(graph_groups, graph_groups_indegree, m, True)
        item_list = self.topo_sort(graph_items, graph_items_indegree,n)
        if group_list == [] or item_list == []: return [] 

        group_to_items = defaultdict(list)
        for item in item_list: 
            group_to_items[group[item]].append(item)
        
        ans = []
        for group_num in group_list:
            ans += group_to_items[group_num]
        
        return ans
        
    def make_graph_groups(self, n, group, beforeItems, graph_groups):
        # seen = set([])
        graph_groups_indegree = defaultdict(int)
        for item in range(len(group)):
            to_group = group[item]
            from_items = beforeItems[item]
            for pre in from_items: 
                from_group = group[pre]
                if to_group == from_group : continue
                graph_groups[from_group].append(to_group)
                graph_groups_indegree[to_group] += 1
                # seen.add((from_group, to_group))
        return graph_groups_indegree

    def make_graph_items(self,n, group, beforeItems, graph_items):
        graph_items_indegree = defaultdict(int)
        for item in range(n):
            before_items = beforeItems[item]
            for pre in before_items: 
                graph_items[pre].append(item)
                graph_items_indegree[item] += 1

        return graph_items_indegree    
    
    def topo_sort(self, graph, indegree, num, printt=False):
        order = []
        q = []

        for node in range(num):
            degree = indegree[node]
            if degree == 0: 
                q.append(node)
                
        while len(q) > 0: 
            node = q.pop(0)
            order.append(node)
            num -= 1 # tracking if we've seen all nodes or not - if cycle, cant visit them all
            for neigh in graph[node]:
                indegree[neigh] -= 1 
                if indegree[neigh] == 0: 
                    q.append(neigh)
            
        return order if num == 0 else []
                

        

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