<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]:        
        indegree = [0 for _ in range(N)]
        adj = defaultdict(list)        
        q = collections.deque()
        r_group_within = defaultdict(list)        

        for i in range(N):
            if group[i] == -1:
                group[i] = M
                M += 1

        g_indegree = [set([]) for _ in range(M)]
        g_adj = defaultdict(set)
        g_q = collections.deque()
        r_group = []

        for i in range(N):
            for elem in beforeItems[i]:
                indegree[i] += 1
                adj[elem].append(i)

                if group[elem] != group[i]:
                    g_indegree[group[i]].add(group[elem])
                    g_adj[group[elem]].add(group[i])

            if indegree[i] == 0:
                q.append(i)

        for g in range(M):
            if len(g_indegree[g]) == 0:
                g_q.append(g)

        while q:
            for _ in range(len(q)):
                item = q.popleft()

                r_group_within[group[item]].append(item)

                for elem in adj[item]:
                    indegree[elem] -= 1

                    if indegree[elem] == 0:
                        q.append(elem)

        # print(g_indegree)
        # print(g_q)
        
        while g_q:
            for _ in range(len(g_q)):
                item = g_q.popleft()

                r_group.append(item)

                for elem in g_adj[item]:
                    g_indegree[elem].remove(item)

                    if len(g_indegree[elem]) == 0:
                        g_q.append(elem)

        # print(indegree)
        # print(g_indegree)
        if any([g for g in g_indegree if len(g) > 0]) or any([g for g in indegree if g > 0]):
            return []

        ans = []

        for g in r_group:
            ans += r_group_within[g]

        return ans
49906