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

        itemIndegrees = [0] * n
        groupIndegrees = [0] * TEMP
        groupEdges = [[] for _ in range(TEMP)]
        for i in range(len(beforeItems)):
            for j in beforeItems[i]:
                itemIndegrees[j] += 1
                if group[i] != group[j]:
                    groupIndegrees[group[j]] += 1
                    # if group[j] not in groupEdges[group[i]]:
                    groupEdges[group[i]].append(group[j])

        # topo sort for items
        queue = []
        for i in range(len(itemIndegrees)):
            if (itemIndegrees[i] == 0):
                queue.append(i)
        sortedItems = []
        while len(queue) > 0:
            I = queue.pop(0)
            sortedItems.append(I)
            for i in beforeItems[I]:
                itemIndegrees[i] -= 1
                if (itemIndegrees[i] == 0):
                    queue.append(i)
        if len(sortedItems) != n:
            return []
            # there is a cycle, impossible
        
        # topo sort for groups
        queue = []
        for i in range(len(groupIndegrees)):
            if (groupIndegrees[i] == 0):
                queue.append(i)
        sortedGroups = []
        while len(queue) > 0:
            I = queue.pop(0)
            sortedGroups.append(I)
            for i in groupEdges[I]:
                groupIndegrees[i] -= 1
                if groupIndegrees[i] == 0:
                    queue.append(i)
        # print(sortedGroups, groupIndegrees)
        if len(sortedGroups) != TEMP:
            return []
            # there is a cycle, impossible

        # merge them.
        sortedItems.reverse()
        sortedGroups.reverse()
        print(sortedItems, sortedGroups)

        groupItems = {}
        for i in sortedItems:
            if group[i] not in groupItems:
                groupItems[group[i]] = []
            groupItems[group[i]].append(i)

        result = []
        for g in sortedGroups:
            if g in groupItems:
                result += groupItems[g]

        return result

446