<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]:
        dct = defaultdict(list) # {group_id: (lvl,val)}
        adj_lst = defaultdict(list)
        income_count = defaultdict(int)
        for i in range(n):
            if group[i] == -1:
                group[i] = m
                m += 1
            income_count[i] = len(beforeItems[i])
            for x in beforeItems[i]:
                adj_lst[x].append(i)
        queue = deque()
        x = n
        for i in range(n):
            if not income_count[i]:
                queue.append(i)
                x -= 1
        lvl = 0
        res = []
        adj_lst2 = defaultdict(list)
        income_count2 = defaultdict(int)
        while queue:
            size = len(queue)
            while size:
                size -= 1
                node = queue.popleft()
                dct[group[node]].append((lvl, node))
                for nei in adj_lst[node]:
                    if group[nei] != group[node]:
                        income_count2[group[nei]] += 1
                        adj_lst2[group[node]].append(group[nei])
                    income_count[nei] -= 1
                    if not income_count[nei]:
                        queue.append(nei)
                        x -= 1
            lvl += 1
        queue = deque()
        for gr in range(m):
            if not income_count2[gr]:
                queue.append(gr)
        if not income_count2[-1]:
            queue.append(-1)
        while queue:
            node = queue.popleft()
            tmp = dct[node]
            for _, v in sorted(tmp):
                res.append(v)
            for nei in adj_lst2[node]:
                income_count2[nei] -= 1
                if not income_count2[nei]:
                    queue.append(nei)

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