<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=defaultdict(list)
        val_to_group=defaultdict(int)
        var=-1
        for i,val in enumerate(group):
            if val==-1:
                groups[var].append(i)
                val_to_group[i]=var
                group[i]=var
                var-=1
                continue
            groups[val].append(i)
            val_to_group[i]=val
        print("val to groups", val_to_group)

        adj_items=defaultdict(list)
        for i,bf in enumerate(beforeItems):
            adj_items[i].extend(bf)

        adj_groups=defaultdict(list)
        for i,val in enumerate(group):
            for each in beforeItems[i]:
                if val_to_group[each]!=val:
                    adj_groups[val].append(val_to_group[each])
        print("groups-adj",adj_groups)


        
        def helper(item,key):
            if item in output:
                return True
            if item in visit:
                return False
            visit.add(item)
            for pre in adj_items[item]:
                if pre not in groups[key]:
                    continue
                if not helper(pre,key):
                    return False
            output.append(item)
            visit.remove(item)
            return True


        for key in groups.keys():
            output=[]
            visit=set()
            for each in groups[key]:
                if each not in output:
                    if not helper(each,key):
                        return []
            groups[key]=output
        print("sorted groups",groups)

        def helper2(grp):
            if grp in visited_groups:
                return True
            if grp in path:
                return False
            path.add(grp)
            for g in adj_groups[grp]:
                if not helper2(g):
                    return False
            visited_groups.append(grp)
            result.extend(groups[grp])
            path.remove(grp)
            return True


        visited_groups=[]
        result=[]
        path=set()
        for g in groups:
            if not helper2(g):
                return []
        return result
            

        
        
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