<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]:
        adj = defaultdict(set)
        inDegree = [0] * n # Indegree of items
        groupInDegree = defaultdict(lambda: 0) # indegree of groups 
        groupDependsOn = defaultdict(list)
        for item, g in enumerate(group):
            adj[g].add(item)
            inDegree[item] = len(beforeItems[item])
            for beforeItem in beforeItems[item]: # Keep track of group dependancy graph without -1 group
                if g != -1 and group[beforeItem] != g and group[beforeItem] != -1: 
                    groupDependsOn[group[beforeItem]].append(g)
                    groupInDegree[g] += 1
        ## Top Sort for groups, without -1 group
        groupOrdering = []
        visited = set()
        q = deque()
        for g in range(m):
            if groupInDegree[g] == 0:
                q.append(g)
        
        while len(q):
            g = q.popleft()
            if g in visited: continue
            visited.add(g)

            for child in groupDependsOn[g]:
                groupInDegree[child] -= 1
                if groupInDegree[child] == 0:
                    q.append(child)
            groupOrdering.append(g)
        print(groupOrdering)

        if len(groupOrdering) != m: return []

        dependsOn = defaultdict(list)
        for item, pres in enumerate(beforeItems):
            if len(pres) == 0: continue
            for pre in pres:
                dependsOn[pre].append(item)
        print(groupInDegree)
        res = []
        visited = set() 
         #Top Sort
        output = []
        q = deque()
        for itemWithNoGroup in adj[-1]:
            if inDegree[itemWithNoGroup] == 0 and itemWithNoGroup not in visited:
                q.append(itemWithNoGroup)
        while len(q):
            item = q.popleft()
            if item in visited: continue

            for child in dependsOn[item]:
                inDegree[child] -= 1
                if inDegree[child] == 0 and group[child] == group[item]:
                    q.append(child)
            
            visited.add(item)
            output.append(item)
        res = res + output
        for g2 in groupOrdering:
            items = adj[g2]
            if len(items) == 0: continue
            #Top Sort
            if g2 != -1:
                output = []
                q = deque()
                for itemWithNoGroup in adj[-1]:
                    if inDegree[itemWithNoGroup] == 0 and itemWithNoGroup not in visited:
                        q.append(itemWithNoGroup)
                while len(q):
                    item = q.popleft()
                    if item in visited: continue

                    for child in dependsOn[item]:
                        inDegree[child] -= 1
                        if inDegree[child] == 0 and group[child] == group[item]:
                            q.append(child)
                    
                    visited.add(item)
                    output.append(item)
                res = res + output  
            output = []
            q = deque()
            for item in items:
                if inDegree[item] == 0 and item not in visited:
                    q.append(item)
            print(g2, q)
            while len(q):
                item = q.popleft()
                if item in visited: continue

                for child in dependsOn[item]:
                    inDegree[child] -= 1
                    if inDegree[child] == 0 and group[child] == group[item]:
                        q.append(child)
                
                visited.add(item)
                output.append(item)
            res = res + output

    #Top Sort
        output = []
        q = deque()
        for itemWithNoGroup in adj[-1]:
            if inDegree[itemWithNoGroup] == 0 and itemWithNoGroup not in visited:
                q.append(itemWithNoGroup)
        while len(q):
            item = q.popleft()
            if item in visited: continue

            for child in dependsOn[item]:
                inDegree[child] -= 1
                if inDegree[child] == 0 and group[child] == group[item]:
                    q.append(child)
            
            visited.add(item)
            output.append(item)
        res = res + output

                    
        print(res)
        return res if len(res) == n else []
45181