<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]:
        
        ggp = defaultdict(set)
        gp = defaultdict(lambda: defaultdict(set))
        gus = defaultdict(set)
        res = []
        gi = -1
        for i in range(len(group)):
            if group[i] == -1:
                group[i] = gi
                gi -= 1
            gus[group[i]].add(i)
        for v, us in enumerate(beforeItems):
            for u in us:
                if group[u] != group[v]:
                    ggp[group[v]].add(group[u])
                else:
                    gp[group[u]][v].add(u)

        def top_sort(us, gp):
            ret = []
            d = defaultdict(int)
            for v in gp:
                for u in gp[v]:
                    d[u] += 1

            stk = [u for u in us if d[u] == 0]
            while stk:
                v = stk.pop()
                ret.append(v)
                for u in gp[v]:
                    d[u] -= 1
                    if d[u] == 0:
                        stk.append(u)
            return [] if len(us) != len(ret) else ret[::-1]

        ggs = top_sort(set(group), ggp)
        # print(ggp)
        # print(gus)
        # print(gp)
        # print(ggs)
        if not ggs:
            return []
        for g in ggs:
            # print(g)
            # print(us)
            # print(gp)
            gs = top_sort(gus[g], gp[g])
            if not gs:
                return []
            res += gs
        return res




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