<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]:
        for i in range(n):
            if group[i] == -1: group[i] = i + m

        in_deg_group = {}
        graph = {}

        unique_groups = set(group)
        for g in unique_groups:
            in_deg_group[g] = set()
            graph[g] = set()

        for i in range(n):
            group_x = group[i]
            for item in beforeItems[i]:
                group_y = group[item]
                if group_x == group_y: continue
                in_deg_group[group_x].add(group_y)
                graph[group_y].add(group_x)

        queue = []
        order = []
        for g in in_deg_group:
            in_deg_group[g] = len(in_deg_group[g])
            if not in_deg_group[g]: queue.append(g)

        while queue:
            cur = queue.pop(0)
            order.append(cur)

            for next_g in graph[cur]:
                in_deg_group[next_g] -= 1
                if not in_deg_group[next_g]: queue.append(next_g)

        if len(order) != len(unique_groups): return []
        

        group_item = {g: [] for g in unique_groups}
        for i in range(n):
            group_item[group[i]].append(i)
        def get_order(group_id):
            in_deg = {i:0 for i in group_item[group_id]}
            graph = {i:[] for i in group_item[group_id]}

            for i in group_item[group_id]:
                for j in beforeItems[i]:
                    if group[i] != group[j]: continue
                    in_deg[i] += 1
                    graph[j].append(i)

            queue = []
            for k, v in in_deg.items():
                if not v: queue.append(k)

            item_order = []
            while queue:
                cur = queue.pop(0)
                item_order.append(cur)

                for next_i in graph[cur]:
                    in_deg[next_i] -= 1
                    if not in_deg[next_i]: queue.append(next_i)

            if len(item_order) != len(group_item[group_id]): return []
            return item_order

        item_order = []
        for g in order:
            group_order = get_order(g)
            if not group_order: return []
            item_order += group_order
        return item_order


        



        
        



        



        
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