<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]:
        


        '''
        n items - each belongs to zero to more groups 
        There are m groups
        A group can have no items 

        - items in same group - next to each other in sorted list 
        - beforeItems - list contains all items that should be before i 
        
        No Group = 0, 1, 7 
        Group 1 = 2, 5 
        Group 0 = 6, 4, 3 

        COME to the left of the ith items ... 

        prereq
        3 ---> 4
        6 ---->
        '''
        def topsort(succ, predec_count, n):
            order = []
            queue = deque(node for node in range(n) if not predec_count[node])
            
            while queue:
                node = queue.popleft()
                order.append(node)
                for s in succ[node]:
                    predec_count[s] -= 1 
                    if not predec_count[s]:
                        queue.append(s)

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

        for item in range(n):
            if group[item] == -1:
                group[item] = m 
                m += 1 
        
        succ_group, succ_item = [[] for _ in range(m)], [[] for _ in range(n)]
        predec_count_group = [0]*m
        predec_count_item = [0]*n

        for item in range(n):
            current = group[item]
            for before in beforeItems[item]:
                before_group = group[before]
                if current == before_group:
                    succ_item[before].append(item)
                    predec_count_item[item] += 1 
                else:
                    succ_group[before_group].append(current)
                    predec_count_group[current] += 1 
        
        grp_order = topsort(succ_group, predec_count_group, m)
        items_order = topsort(succ_item, predec_count_item, n)

        if not grp_order or not items_order:
            return []
        
        items_grouped = [[] for _ in range(m)]
        for item in items_order:
            items_grouped[group[item]].append(item)

        ans = []
        for g in grp_order:
            ans.extend(items_grouped[g])
        
        return ans
33368