<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>

class Solution:

    def __topoSort(self, adj: defaultdict[int, list[int]], indegree: list[int], n: int) -> list[int]:
        tpSort: list[int] = list()
        dq: deque[int] = deque()

        for node, degree in enumerate(indegree):
            if not degree:
                dq.append(node)
        
        while dq:
            node = dq.popleft()
            tpSort.append(node)

            for nei in adj[node]:
                indegree[nei] -= 1
                if not indegree[nei]:
                    dq.append(nei)

        return tpSort if len(tpSort) == n else []

    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        groups: defaultdict[int, list[int]] = defaultdict(list)
        items: defaultdict[int, list[int]] = defaultdict(list)
        indegreeGrp: list[int] = [0]*m
        indegreeItm: list[int] = [0]*n
        grpIndex: defaultdict[int, list[int]] = defaultdict(list)
        ans: list[int] = list()

        for item, grp in enumerate(group):
            if grp == -1: 
                group[item], grp, m = m, m, m+1
                indegreeGrp.append(0)

            for bfrItem in beforeItems[item]:
                if group[bfrItem] == -1: 
                    group[bfrItem], m = m, m+1
                    indegreeGrp.append(0)

                if group[bfrItem] != grp:
                    groups[group[bfrItem]].append(grp) 
                    indegreeGrp[grp] += 1
                
                items[bfrItem].append(item)
                indegreeItm[item] += 1
        
        grpOrder = self.__topoSort(groups, indegreeGrp, m)
        itmOrder = self.__topoSort(items, indegreeItm, n)
        if not itmOrder or not grpOrder:
            return []
        
        for item in itmOrder:
            grpIndex[group[item]].append(item)

        for grp in grpOrder:
            ans.extend(grpIndex[grp])

        return ans
38431