<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]:
        groups = {}
        itemToNext = {}
        itemToPrev = {}
        groupToNext = {}
        groupToPrev = {}
        newGroup = m+1
        for i in range(n):
            if group[i] != -1:
                if group[i] not in groups:
                    groups[group[i]] = set()
                groups[group[i]].add(i)
            else:
                groups[newGroup] = set()
                groups[newGroup].add(i)
                group[i]=newGroup
                newGroup +=1
            if len(beforeItems[i])>0:
                prevs = beforeItems[i]
                for prev in prevs:
                    if prev not in itemToNext:
                        itemToNext[prev] = set()
                    itemToNext[prev].add(i)
                    if i not in itemToPrev:
                        itemToPrev[i] = set()
                    itemToPrev[i].add(prev)
        for i in range(n):
            if len(beforeItems[i])>0:
                prevs = beforeItems[i]
                for prev in prevs:
                    if group[prev] != group[i]:
                        if group[prev] not in groupToNext:
                            groupToNext[group[prev]] = set()
                        groupToNext[group[prev]].add(group[i])
                    if group[prev] != group[i]:
                        if group[i] not in groupToPrev:
                            groupToPrev[group[i]] = set()
                        groupToPrev[group[i]].add(group[prev]) 
        answer = []
        prevGroup = -1
        while(len(groups)>0):
            found = False
            if prevGroup == -1:
                for g in groups:
                    if not found:
                        for item in groups[g]:
                            if g not in groupToPrev or len(groupToPrev[g])== 0:
                                if item not in itemToPrev or len(itemToPrev[item])==0:
                                    found = True
                                    break
            else:
                for item in groups[prevGroup]:
                    if item not in itemToPrev or len(itemToPrev[item])==0:
                        found = True
                        break

            if not found:
                return []
            answer.append(item)
            if item in itemToNext:
                nexts = itemToNext[item]
                for n in nexts:
                    itemToPrev[n].remove(item)
            g = group[item]
            groups[g].remove(item)
            if len(groups[g])==0:
                groups.pop(g)
                prevGroup = -1
                if g in groupToNext:
                    nextGroups = groupToNext[g]
                    for ng in nextGroups:
                        groupToPrev[ng].remove(g)
            else:
                prevGroup = g
        return answer


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