<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 DirectedGraph:
    def __init__(self):
        self.graph: Dict[str, List[int]] = defaultdict(list)
        self.indegree: Dict[str, int] = defaultdict(int)
        self.nodes: Set[int] = set()

    def add_edge(self, from_node: int, to_node: int):
        self.graph[from_node].append(to_node)
        self.indegree[to_node] += 1
        self.nodes.add(from_node)
        self.nodes.add(to_node)

    def size(self) -> int:
        return len(self.nodes)

    def get_order(self) -> List[int]:
        ans = []

        queue = deque()

        for node in self.nodes:
            if self.indegree[node] == 0:
                queue.append(node)

        while queue:
            out = queue.pop()
            ans.append(out)

            for neighbor in self.graph[out]:
                self.indegree[neighbor] -= 1

                if self.indegree[neighbor] == 0:
                    queue.appendleft(neighbor)

        return ans


class Solution:
    def sortItems(
        self, n: int, m: int, group: List[int], beforeItems: List[List[int]]
    ) -> List[int]:
        group_ids = []
        for i in range(n):
            if group[i] == -1:
                group_ids.append("g_" + str(i))
            else:
                group_ids.append(str(group[i]))

        group_items: Dict[str, Set[int]] = defaultdict(set)

        group_graph = DirectedGraph()
        internal_graphs: Dict[str, DirectedGraph] = defaultdict(DirectedGraph)

        for i in range(n):
            group_id = group_ids[i]
            group_items[group_id].add(i)

            for before in beforeItems[i]:
                before_item_group_id = group_ids[before]

                if group_id != before_item_group_id:
                    group_graph.add_edge(before_item_group_id, group_id)
                else:
                    internal_graphs[group_id].add_edge(before, i)

        result = []
        group_order = group_graph.get_order()
        group_set = set(group_order)

        if len(group_order) != group_graph.size():
            return result

        for group_id in group_items:
            if group_id not in group_set:
                group_order.append(group_id)

        for group_id in group_order:
            internal_order = internal_graphs[group_id].get_order()

            if len(internal_order) != internal_graphs[group_id].size():
                return []

            remaining = group_items[group_id] - set(internal_order)

            internal_order.extend(list(remaining))
            result.extend(internal_order)

        return result

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