<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]:
        # topological sorting

        # first, we sort the group, then we sort within each group, and then merge the two results

        # renumber the group such that -1 has new unique group number
        k = m
        for i in range(n):
            if group[i] == -1:
                group[i] = k
                k += 1
        # print("Group after renumbering", group)
        
        # get the prerequisites relation between groups
        group_prerequisites = {i: [] for i in range(k)}
        within_group_prerequisites = {i: {} for i in range(k)}
        in_degrees_group = [0] * k
        in_degrees_within_group = {i: {} for i in range(k)}

        for i in range(n):
            within_group_prerequisites[group[i]][i] = []
            in_degrees_within_group[group[i]][i] = 0

        for i in range(n):
            for pre in beforeItems[i]:
                if group[pre] != group[i]:
                    in_degrees_group[group[i]] += 1
                    group_prerequisites[group[pre]].append(group[i])
                else:
                    within_group_prerequisites[group[pre]][pre].append(i)
                    in_degrees_within_group[group[pre]][i] += 1
        # print("prerequisites between groups",group_prerequisites)
        # print("within each group", within_group_prerequisites)


        # sort the group by topological sorting
        sorted_group = []
        for i in range(k):
            if in_degrees_group[i] == 0:
                sorted_group.append(i)
        
        visited_group = sorted_group.copy()
        while visited_group:
            kk = len(visited_group)
            for _ in range(kk):
                curr = visited_group.pop(0)
                for nxt in group_prerequisites[curr]:
                    in_degrees_group[nxt] -= 1
                    if in_degrees_group[nxt] == 0:
                        sorted_group.append(nxt)
                        visited_group.append(nxt)
        if len(sorted_group) != k:
            return []
        
        # now, sort the items within each group
        sorted_within_group = {i: [] for i in range(k)}
        for i in range(k):
            sorted_curr_group = []
            for j in in_degrees_within_group[i]:
                if in_degrees_within_group[i][j] == 0:
                    sorted_curr_group.append(j)
            visited = sorted_curr_group.copy()

            while visited:
                kk = len(visited)
                for _ in range(kk):
                    curr = visited.pop(0)
                    for nxt in within_group_prerequisites[i][curr]:
                        in_degrees_within_group[i][nxt] -= 1
                        if in_degrees_within_group[i][nxt] == 0:
                            sorted_curr_group.append(nxt)
                            visited.append(nxt)
            sorted_within_group[i] = sorted_curr_group.copy()

        res = []
        for i in sorted_group:
            res += sorted_within_group[i]
        return res if len(res) == n else []


        
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