<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]:
        '''
        first do a topological sort of just groups
        
        then do a topological sort inside each group
        
        '''
        
        def sort(adj_list):
            
            rec = set()
            vis = set()
            
            res = []
            
            def find(u):
                if u in vis:
                    return True
                
                rec.add(u)
                vis.add(u)
                
                for v in adj_list[u]:
                    if v in rec or not find(v):
                        return False
                    
                res.append(u)
                rec.remove(u)
                return True
            
            for u in list(adj_list.keys()):
                if not find(u):
                    return None
            
            return res[::-1]
        
        group_list = defaultdict(list)
        
        same_group_list = [defaultdict(list) for _ in range(m)]
        
        group_mapper = defaultdict(list)
        
        for i in range(n):
            if group[i] == -1:
                group[i] = i + m
            
            group_mapper[group[i]].append(i)
        
        for i, arr in enumerate(beforeItems):
            for j in arr:                    
                if group[i] != group[j]:
                    group_list[group[j]].append(group[i])
                    
                else:
                    same_group_list[group[j]][j].append(i)
        
        sorted_groups = sort(group_list)
        
        if group_list and not sorted_groups:
            return []
        
        sorted_inside_groups = defaultdict(list)
        
        for group, d in enumerate(same_group_list):
            if not d:
                continue
                
            sorted_inside_groups[group] = sort(d) 
            if not sorted_inside_groups[group]:
                return []
        
        done = [False] * n
        
        res = []
        # print(group_mapper)
        # print(sorted_groups)
        # print(sorted_inside_groups)
        
        for x in sorted_groups:
            for item in sorted_inside_groups[x]:
                res.append(item)
                done[item] = True
                
            for item in group_mapper[x]:
                if not done[item]:
                    res.append(item)
                    done[item] = True
        
        for i in range(m + n):
            for item in sorted_inside_groups[i]:
                if not done[item]:
                    res.append(item)
                    done[item] = True
                    
            for item in group_mapper[i]:
                if not done[item]:
                    res.append(item)
                    done[item] = True
    
 
        
        return res
            
            
            
            
        
        
        
        
                    
        
        
        
                    
                    
                
        
        
47543