<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]:
        groups = collections.defaultdict(list)
        get_group = lambda x: group[x] if (group[x] != -1) else m + x
        group_graph = set()

        adj_mat = [[] for _ in range(n)]
        in_vertices = [0] * n

        for i, items in enumerate(beforeItems):
            for j in items:
                adj_mat[j].append(i)
                in_vertices[i] += 1
        
        queue = collections.deque([x for x in range(n) if in_vertices[x] == 0])
        while(queue):
            curr = queue.popleft()
            groups[get_group(curr)].append(curr)

            for next in adj_mat[curr]:
                p_group = get_group(curr)
                c_group = get_group(next)
                if(p_group != c_group): group_graph.add((p_group, c_group))
                in_vertices[next] -= 1
                if(in_vertices[next] == 0):
                    queue.append(next)
        # print(group_graph)
        
        if(any([x for x in range(n) if in_vertices[x] != 0])): return []

        g_adj_mat = collections.defaultdict(list)
        g_in_vertices = collections.defaultdict(int)

        for x,y in group_graph:
            g_adj_mat[x].append(y)
            g_in_vertices[y] += 1
        
        queue = collections.deque([x for x in range(m+n) if g_in_vertices[x] == 0])
        output = []
        while(queue):
            curr = queue.popleft()
            for item in groups[curr]:
                output.append(item)
            
            for next in g_adj_mat[curr]:
                g_in_vertices[next] -= 1
                if(g_in_vertices[next] == 0):
                    queue.append(next)
        
        if(any([x for x in range(m + n) if g_in_vertices[x] != 0])): return []
        return output


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