<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>

#Assign group to each item.
#Create connection between the groups.
#create connection of items within each group.
#Topological sort of group.
#Within group perform topological sort of items.
class Solution:
    def sortItems(self, n: int, m: int, group: [int], beforeItems: [[int]]) -> [int]:
        groupId = {}
        groupNum = 1
        itemToGroupMap = {}
        groupToItem = defaultdict(list)
        # assign gourpnum to each item.
        for i in range(0, len(group)):
            if group[i] == -1:
                itemToGroupMap[i] = groupNum
                groupToItem[groupNum].append(i)
                groupNum += 1
            else:
                if group[i] in groupId:
                    gId = groupId[group[i]]
                    itemToGroupMap[i] = gId
                    groupToItem[gId].append(i)
                else:
                    groupId[group[i]] = groupNum
                    groupToItem[groupNum].append(i)
                    groupNum += 1
                itemToGroupMap[i] = groupId[group[i]]

        groupG = defaultdict(list)
        itemGraph = {}  # {item: defaultdict(list)}
        for i in range(0, len(beforeItems)):
            if len(beforeItems[i]) != 0:
                for v in beforeItems[i]:
                    groupId1 = itemToGroupMap[i]
                    groupId2 = itemToGroupMap[v]
                    if groupId1 == groupId2:
                        # add graph in individual group
                        if groupId1 in itemGraph:
                            g = itemGraph[groupId1]
                        else:
                            g = defaultdict(list)
                        g[i].append(v)
                        itemGraph[groupId1] = g
                    else:
                        groupG[groupId1].append(groupId2)

        # perform topological sort on group.
        if  self.is_cyclic_directed(groupG):
            return []

        groupOrder = self.topoSortUtilGroup(groupG, groupNum)
        print(f"GroupOrder = {groupOrder}")
        res = []
        while len(groupOrder) != 0:
            # check if this group has graph.
            groupId = groupOrder.pop(0)
            if groupId in itemGraph:
                G = itemGraph[groupId]
                if self.is_cyclic_directed(G):
                    return []
                res1 = self.topoSortUtil(G)
                for item in groupToItem[groupId]:
                    if not item in res1:
                        res1.append(item)
                res += res1
            else:
                # add items of groupId
                res += groupToItem[groupId]
        return res

    def topoSortUtil(self, G):
        visited = set()
        stack = []
        for node in G:
            if not node in visited:
                self.topoSort(node, G, stack, visited)
        return stack

    def topoSortUtilGroup(self, G, n):
        visited = set()
        stack = []
        for node in range(1, n):
            if not node in visited:
                self.topoSort(node, G, stack, visited)
        return stack

    def topoSort(self, node, G, L, visited):
        visited.add(node)
        if node in G:
            for v in G[node]:
                if not v in visited:
                    self.topoSort(v, G, L, visited)
        L.append(node)

    def is_cyclic_directed_util(self,v, visited, rec_stack, graph):
        # Mark the current node as visited and add it to the recursion stack
        visited.add(v)
        rec_stack.append(v)

        # Recur for all neighbors
        if v in graph:
            for neighbor in graph[v]:
                # If the neighbor is not visited, then recur on it
                if not neighbor in visited:
                    if self.is_cyclic_directed_util(neighbor, visited, rec_stack, graph):
                        return True
                # If the neighbor is in the recursion stack, a cycle is detected
                elif neighbor in rec_stack:
                    return True

        # Remove the vertex from the recursion stack
        rec_stack.remove(v)
        return False

    def is_cyclic_directed(self, graph):
        num_vertices = len(graph)
        visited = set()
        rec_stack = []

        # Call the recursive helper function to detect cycle in different DFS trees
        for node in graph:
            if not node in visited:
                if self.is_cyclic_directed_util(node, visited, rec_stack, graph):
                    return True
        return False

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