<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>

import copy
from collections import deque

class Solution:
    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        #1. The items that belong to the same group are next to each other in the sorted list.
        #2. There are some relations between these items where beforeItems[i] is a list containing all the items that should come before the i-th item in the sorted array (to the left of the i-th item).

        #So it is toplogical sort with constraint 1. 

        #Let's first write toplogical sort and then make the 
        group2items = {}
        item2group={}
        counter = m
        for gi, g in enumerate(group):
            k = g
            if g==-1:
                k = counter
                group2items[counter] = set()        
                counter +=1    
            elif not(g in group2items):
                group2items[g] = set()
            group2items[k].add(gi)
            item2group[gi] = k

        #Make group adjacency list 
        group_adjacency_list = {}
        for g in group2items:
            group_adjacency_list[g] = set()

        #Make item adjacency list
        item_adjacency_list = {}
        for item in range(n):
            item_adjacency_list[item] = set()

        for item_after, items in enumerate(beforeItems):
            for item in items:
                item_adjacency_list[item].add(item_after)     
                #Do group adjacency list 
                if item2group[item] != item2group[item_after]:
                    group_adjacency_list[item2group[item]].add(item2group[item_after])
        
        visited = set()
        path = set()
        top_sort = []

        def dfs(item, adjacency_list, use_group=None):
            if use_group!= None and item2group[item] != use_group:
                return True
            #check if in path 
            if item in path:
                return False
            if item in visited:
                return True
            
            visited.add(item)
            path.add(item)

            for neighbor in adjacency_list[item]:
                result = dfs(neighbor, adjacency_list, use_group)
                if result ==False:
                    return False
            
            path.remove(item)
            top_sort.append(item)
            return True
            

        #Do topsort among the groups
        visited = set()
        path = set()
        top_sort = []
        for group in group2items:    
            result =dfs(group, group_adjacency_list)
            if result == False:
                return []
        group_top_sort = copy.deepcopy(top_sort[::-1])
        # #Do topsort among the items in the same group 
        # for group in group_top_sort:
        #     #sort inside each group 

        #Do top sort with item adjacency list 
        item_top_sort = []
        for g in group_top_sort:
            visited = set()
            path = set()
            top_sort = []
            # g_item_adjacency_list = {}
            # for k, v in item_adjacency_list.items():
            #     if k in group2items[g]:
            #         g_item_adjacency_list[k] = set(vi for vi in v if vi in group2items[g])
            for item in group2items[g]:    
                #result=dfs(item, g_item_adjacency_list)
                result=dfs(item, item_adjacency_list, use_group = g)
                if result == False:
                    return []
            item_top_sort.extend(top_sort[::-1])
            #item_top_sort.extend(copy.deepcopy(top_sort[::-1]))
            # print("g", g)
            # print("top_sort[::-1]", top_sort[::-1])
        return item_top_sort
        
# import copy
# from collections import deque

# class Solution:
#     def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
#         group2items = {}
#         item2group = {}
#         counter = m
#         for gi, g in enumerate(group):
#             if g == -1:
#                 k = counter
#                 group2items[counter] = set()        
#                 counter += 1                
#             else:
#                 k = g 
#                 group2items.setdefault(k, set())
#             group2items[k].add(gi)
#             item2group[gi] = k
        
#         # Initialize group adjacency list correctly
#         group_adjacency_list = {g: set() for g in group2items}

#         # Make item adjacency list
#         item_adjacency_list = {i: set() for i in range(n)}

#         for item_after, items in enumerate(beforeItems):
#             for item in items:
#                 item_adjacency_list[item].add(item_after)
#                 if item2group[item] != item2group[item_after]:
#                     group_adjacency_list[item2group[item]].add(item2group[item_after])
        
#         def dfs(item, adjacency_list):
#             if item in path:
#                 return False  # Cycle detected
#             if item in visited:
#                 return True  # Already visited safely

#             visited.add(item)
#             path.add(item)

#             for neighbor in adjacency_list[item]:
#                 if not dfs(neighbor, adjacency_list):
#                     return False

#             path.remove(item)
#             top_sort.append(item)
#             return True
        
#         # Do topsort among the groups
#         visited = set()
#         path = set()
#         top_sort = []
#         for group in group2items:    
#             if not dfs(group, group_adjacency_list):
#                 return []
#         group_top_sort = copy.deepcopy(top_sort[::-1])

#         # Do top sort with item adjacency list 
#         item_top_sort = []
#         for g in group_top_sort:
#             g_item_adjacency_list = {item: set() for item in group2items[g]}
#             for k, v in item_adjacency_list.items():
#                 if k in group2items[g]:
#                     g_item_adjacency_list[k] = v.intersection(group2items[g])

#             visited = set()
#             path = set()
#             top_sort = []
#             for item in group2items[g]:    
#                 if not dfs(item, g_item_adjacency_list):
#                     return []
#             item_top_sort.extend(top_sort[::-1])
#             print("g", g)
#             print("top_sort[::-1]", top_sort[::-1])

#         return item_top_sort


        
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