<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)):
            num = group[i]
            if num == -1:
                group[i] = m
                m += 1
        # build edges, group lists
        groupdict = defaultdict(list)
        for i in range(len(group)):
            num = group[i]
            groupdict[num].append(i)

        intraEdges = defaultdict(list)
        interEdges = defaultdict(list)
        item_degree = [0 for _ in range(n)]
        group_degree = [0 for _ in range(m)]

        for i in range(len(beforeItems)):
            for node in beforeItems[i]:
                if (group[node] == group[i]):
                    interEdges[node].append(i)
                    item_degree[i] += 1
                else:
                    if group[i] not in intraEdges[group[node]]:
                        intraEdges[group[node]].append(group[i])
                        group_degree[group[i]] += 1

        # for intra group edges, do the topological sort on groups
        def top_kahns(indegree, edges, node):
            queue = []
            output = []
            for i in node:
                if indegree[i] == 0:
                    queue.append(i)
            while queue:
                groupnum = queue.pop(0)
                output.append(groupnum)

                for item in edges[groupnum]:
                    indegree[item] -= 1
                    if indegree[item] == 0:
                        queue.append(item)
            if len(output) < len(node):
                return []
            return output

        group_top = top_kahns(group_degree, intraEdges, list(range(m)))
        if not group_top:
            return []
        
        # Then, sort the items within each group
        result = []
        for groupnum in group_top:
            if groupdict[groupnum]:
                top_eachgroup = top_kahns(item_degree, interEdges, groupdict[groupnum])
                if not top_eachgroup:
                    return []
                result.extend(top_eachgroup)
        
        return result


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