<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 sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        group_g = {}
        item_g = {}
        groupMap = {}

        group_idx = m
        for i in range(len(group)):
            if group[i] == -1:
                groupMap[i] = group_idx
                group_idx += 1
            else: 
                groupMap[i] = group[i]

        groups = {k:[] for k in range(group_idx)}
        for item in groupMap: 
            if groupMap[item] not in groups: 
                groups[groupMap[item]] = []
            groups[groupMap[item]].append(item)

        
        group_dep = {k: set() for k in range(group_idx)}
        item_dep = { k: set() for k in range(n)}

        for i in range (len(beforeItems)): 
            for pre in beforeItems[i]:
                item_dep[i].add(pre)
                if groupMap[i] != groupMap[pre]:
                    group_dep[groupMap[i]].add(groupMap[pre])

        def topoSort(item, graph, visited, cycle, res):
            if item in visited: 
                return True 
            if item in cycle: 
                return False 

            cycle.add(item)
            if item in graph: 
                for n in graph[item]:
                    if not topoSort(n, graph, visited, cycle, res):
                        return False
            
            cycle.remove(item)
            visited.add(item)
            res.append(item)
            return True 
        
        group_order = []
        visited, cycle = set(), set()
        for i in range(group_idx):
            if not topoSort(i, group_dep, visited, cycle, group_order):
                return []

        final_order = []
        visited, cycle = set(), set() 
        for group_num in group_order:
            for item in groups[group_num]:
                if not topoSort(item, item_dep, visited, cycle, final_order):
                    return []
                    
        return final_order
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