<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 has_cycle(self, graph, cur_node, visited, result):
        if visited[cur_node] == 1:
            return False
        if visited[cur_node] == 2:
            return True
        visited[cur_node] = 2
        for next_node in graph[cur_node]:
            if self.has_cycle(graph, next_node, visited, result):
                return True
        visited[cur_node] = 1
        result.append(cur_node)
        return False

    def sortItems(self, n: int, m: int, group: List[int],
                  beforeItems: List[List[int]]) -> List[int]:
        # Map between group_id and item_ids
        group_items_map = defaultdict(list)
        # Visited for items in each group. Will be used later
        visited_item = defaultdict(dict)
        for i in range(n):
            # Assign no-group items to a new group
            if group[i] == -1:
                group[i] = m
                m += 1
            group_items_map[group[i]].append(i)
            visited_item[group[i]][i] = 0

        # key - group_id : value - next_groups
        graph_group = defaultdict(set)
        # key - group_id : value - {key - item_id : value: next_items}
        graph_item = {i: defaultdict(list) for i in range(m)}

        # Create graph for items and groups
        for item_after, before_items in enumerate(beforeItems):
            for item_before in before_items:
                group_before = group[item_before]
                group_after = group[item_after]

                # If two items belong to different groups,
                #   add a dependency between groups
                # Otherwise, add a dependency between items in the same group
                if group_before != group_after:
                    graph_group[group_before].add(group_after)
                else:
                    graph_item[group_before][item_before].append(item_after)

        # Use DFS to find group order
        visited_group = [0] * m
        group_order = []
        for group_id in range(m):
            if self.has_cycle(graph_group, group_id,
                              visited_group, group_order):
                return []

        # Use DFS to find item order in each group
        full_item_order = []
        for group_id in group_order:
            for item_id in group_items_map[group_id]:
                if self.has_cycle(graph_item[group_id], item_id,
                                  visited_item[group_id], full_item_order):
                    return []
        return full_item_order[::-1]


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