<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 build_graph(self, n, m, group, group_id, beforeItems):
        # Sort all item regardless of group dependencies.
        item_graph = [[] for _ in range(n)]
        item_indegree = [0] * n

        # Sort all groups regardless of item dependencies.
        group_graph = [[] for _ in range(group_id)]
        group_indegree = [0] * group_id

        for curr in range(n):
            for prev in beforeItems[curr]:
                # Each (prev -> curr) represents an edge in the item graph.
                item_graph[prev].append(curr)
                item_indegree[curr] += 1
                
                # If they belong to different groups, add an edge in the group graph.
                if group[curr] != group[prev]:
                    group_graph[group[prev]].append(group[curr])
                    group_indegree[group[curr]] += 1  
        
        return group_graph, group_indegree, item_graph, item_indegree
    
    def bfs(self, graph, in_degree):
        # 根据入度初始化队列中的节点
        q = []
        for i in range(len(in_degree)):
            if in_degree[i] == 0:
                q.append(i)

        topological_sort = []
        # 开始执行 BFS 算法
        while q:
            cur = q.pop(0)
            # 弹出节点的顺序即为拓扑排序结果
            topological_sort.append(cur)
            for nei in graph[cur]:
                in_degree[nei] -= 1
                if in_degree[nei] == 0:
                    q.append(nei)

        return topological_sort if len(topological_sort) == len(graph) else []

    def sortItems(self, n: int, m: int, group: List[int], beforeItems: List[List[int]]) -> List[int]:
        # If an item belongs to zero group, assign it a unique group id.
        group_id = m
        for i in range(n):
            if group[i] == -1:
                group[i] = group_id
                group_id += 1
        
        group_graph, group_indegree, item_graph, item_indegree = self.build_graph(
            n, m, group, group_id, beforeItems
        )

        item_order = self.bfs(item_graph, item_indegree)
        group_order = self.bfs(group_graph, group_indegree)

        if not item_order or not group_order: 
            return []
        
        # Items are sorted regardless of groups, we need to 
        # differentiate them by the groups they belong to.
        ordered_groups = collections.defaultdict(list)
        for item in item_order:
            ordered_groups[group[item]].append(item)
        
        # Concatenate sorted items in all sorted groups.
        # [group 1, group 2, ... ] -> [(item 1, item 2, ...), (item 1, item 2, ...), ...]
        answer = []
        for group_index in group_order:
            answer += ordered_groups[group_index]
        return answer



        
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