<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>

"""
Intution

Do Toposort on groups and Items


"""
from collections import deque

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:

        def toposort(adjList : List[List[int]], indegree: List[int],n) -> List[int]:

            queue = deque()

            visited_count = 0

            for i in range(n):

                if indegree[i] == 0:
                    queue.append((i))
            ans = []
            while queue:

                node = queue.popleft()
                visited_count += 1

                if visited_count > n:
                    return []


                ans.append(node)

                for neighbour in adjList[node]:

                    indegree[neighbour] -= 1
                    if indegree[neighbour] == 0:
                        queue.append((neighbour))
            
            if visited_count == n:
                return ans
            
            else:
                return []


        for  i in range(n):

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

        items_indegree = [0]*n
        groups_indegree = [0]*m



        for i in range(n):

            for parent in beforeItems[i]:

                items_adjList[parent].append(i)
                items_indegree[i] += 1

                if group[parent] != group[i]:

                    groups_adjList[group[parent]].append(group[i])
                    groups_indegree[group[i]] += 1
        
        group_toposort = toposort(groups_adjList, groups_indegree, m)
        items_toposort = toposort(items_adjList, items_indegree, n)

        if not group_toposort or not items_toposort:
            return []
        
        ans = []

        items_grouped = [[] for _ in range(m)]

        for item in items_toposort:

            items_grouped[group[item]].append(item)

        for grp in group_toposort:

            ans.extend(items_grouped[grp])

        return ans
        
    


            

            
            
        
                
        
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