<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(n):
            if group[i]==-1:
                group[i]=m
                m+=1
        itemGraph=[[] for _ in range(n)]
        itemIndegree=[0]*n
        groupGraph=[[] for _ in range(m)]
        groupIndegree=[0]*m
        for curr in range(n):
            for prev in beforeItems[curr]:
                itemGraph[prev].append(curr)
                itemIndegree[curr]+=1
                if group[curr]!=group[prev]:
                    prevGroup = group[prev]
                    currGroup = group[curr]
                    groupGraph[prevGroup].append(currGroup)
                    groupIndegree[currGroup] += 1
        itemOrder=self.topologicalSort(itemGraph, itemIndegree,n)
        groupOrder=self.topologicalSort(groupGraph, groupIndegree,m)
        if not itemOrder or not groupOrder:
            return []
        result=[]
        group_to_items = defaultdict(list)
        for item in itemOrder:
            group_id=group[item]
            group_to_items[group_id].append(item)
        result=[]
        for grp in groupOrder:
            result.extend(group_to_items.get(grp, []))
        return result
    def topologicalSort(self, graph: List[List[int]], indegree: List[int], total_nodes: int) -> List[int]:
        # Initialize queue with nodes having indegree 0
        queue = deque([node for node in range(total_nodes) if indegree[node] == 0])
        order = []

        while queue:
            node = queue.popleft()
            order.append(node)
            for neighbor in graph[node]:
                indegree[neighbor] -= 1
                if indegree[neighbor] == 0:
                    queue.append(neighbor)

        # Check if topological sort is possible
        if len(order) == total_nodes:
            return order
        else:
            return []






            


        
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