<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]:
        def bfs(to, nodes):
            dg = defaultdict(int)
            for s in nodes:
                for e in to[s]:
                    dg[e] += 1
            dq = deque([n for n in nodes if dg[n] == 0])
            vi = []
            while dq:
                u = dq.popleft()
                vi.append(u)
                for v in to[u]:
                    dg[v] -= 1
                    if dg[v] == 0:
                        dq.append(v)
            return vi[::-1] if len(vi) == len(nodes) else []

        mp = defaultdict(list)
        for i, g in enumerate(group):
            if g != -1:
                mp[g].append(i)
        gs = {i: i for i in range(n)}  # ufと同様最初は自分自身のみの1要素グループ
        for _, vs in mp.items():
            for v in vs:
                gs[v] = vs[0]  # root idに所属するように設定
        into = defaultdict(list)
        goto = defaultdict(list)
        for i, items in enumerate(beforeItems):
            for j in items:
                if gs[i] == gs[j]:
                    into[i].append(j)  # j <- i
                else:
                    goto[gs[i]].append(gs[j])                    
        goto_rank = bfs(goto, list(set(gs.values())))  # root nodeだけ
        into_rank = bfs(into, list(range(n)))
        if not goto_rank or not into_rank:
            return []
        gorder = defaultdict(list)  # keyはroot nodeのみ
        for i in into_rank:
            gorder[gs[i]].append(i)
        ans = []
        for i in goto_rank:
            ans += gorder[i]
        return ans

        
39106