<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 queue import PriorityQueue
import heapq
from collections import defaultdict
class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        inDegree = [0 for i in range(n)]
        edges = [[] for i in range(n)]
        
        group_id = m
        for i, x in enumerate(group):
            if x == -1:
                group[i] = group_id
                group_id += 1

        group_indegree = [0]*group_id
        groups = [[] for i in range(group_id)]


        for i, rev_edge in enumerate(beforeItems):            
            for j in rev_edge:
                edges[j].append(i)
                inDegree[i] += 1
                if group[i] != group[j]:
                    groups[group[j]].append(group[i])
                    group_indegree[group[i]] += 1
        


        visited = [False]*n
        # print(edges)
        def topological_sort(graph, indegree):
            visited = []
            stack = [node for node in range(len(graph)) if indegree[node] == 0]
            while(stack):
                cur = stack.pop()
                visited.append(cur)
                for neib in graph[cur]:
                    indegree[neib] -= 1
                    if indegree[neib] == 0:
                        stack.append(neib)
            return visited if len(visited)== len(graph) else []

        item_order = topological_sort(edges, inDegree)
        print(item_order)
        group_order = topological_sort(groups, group_indegree)
        print(group_order)
        if not item_order or not group_order:
            return []
        
        group_sort = defaultdict(list)

        for i in item_order:
            group_sort[group[i]].append(i)
        print(group_sort)


        # print(edges, inDegree)
        answer = []
        for group in group_order:
            answer += group_sort[group]
        


        return answer


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