Python Program For Pointing To Next Higher Value Node In A Linked List With An Arbitrary Pointer
Given singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. Need to make the “arbitrary” pointer point to the next higher value node.
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A Simple Solution is to traverse all nodes one by one, for every node, find the node which has the next greater value of the current node and change the next pointer. Time Complexity of this solution is O(n2).
An Efficient Solution works in O(nLogn) time. The idea is to use Merge Sort for linked list.
1) Traverse input list and copy next pointer to arbit pointer for every node.
2) Do Merge Sort for the linked list formed by arbit pointers.
Below is the implementation of the above idea. All of the merger sort functions are taken from here. The taken functions are modified here so that they work on arbit pointers instead of next pointers.
Python3
# Python3 program to populate arbit pointers # to next higher value using merge sort head = None # Link l node class Node: def __init__( self , data): self .data = data self . next = None self .arbit = None # Utility function to print result # linked l def printList(node, anode): print ( "Traversal using Next Pointer" ) while (node ! = None ): print (node.data, end = ", " ) node = node. next print ( "Traversal using Arbit Pointer" ); while (anode ! = None ): print (anode.data, end = ", " ) anode = anode.arbit # This function populates arbit pointer in # every node to the next higher value. And # returns pointer to the node with minimum # value def populateArbit(start): temp = start # Copy next pointers to arbit pointers while (temp ! = None ): temp.arbit = temp. next temp = temp. next # Do merge sort for arbitrary pointers and # return head of arbitrary pointer linked l return MergeSort(start) # Sorts the linked l formed by arbit pointers # (does not change next pointer or data) def MergeSort(start): # Base case -- length 0 or 1 if (start = = None or start.arbit = = None ): return start # Split head into 'middle' and # 'nextofmiddle' sublists middle = getMiddle(start) nextofmiddle = middle.arbit middle.arbit = None # Recursively sort the sublists left = MergeSort(start) right = MergeSort(nextofmiddle) # answer = merge the two sorted lists together sortedlist = SortedMerge(left, right) return sortedlist # Utility function to get the # middle of the linked l def getMiddle(source): # Base case if (source = = None ): return source fastptr = source.arbit slowptr = source # Move fastptr by two and slow ptr by one # Finally slowptr will point to middle node while (fastptr ! = None ): fastptr = fastptr.arbit if (fastptr ! = None ): slowptr = slowptr.arbit fastptr = fastptr.arbit return slowptr def SortedMerge(a, b): result = None # Base cases if (a = = None ): return b elif (b = = None ): return a # Pick either a or b, and recur if (a.data < = b.data): result = a result.arbit = SortedMerge(a.arbit, b) else : result = b result.arbit = SortedMerge(a, b.arbit) return result # Driver code if __name__ = = '__main__' : # Let us create the l shown above head = Node( 5 ) head. next = Node( 10 ) head. next . next = Node( 2 ) head. next . next . next = Node( 3 ) # Sort the above created Linked List ahead = populateArbit(head) print ( "Result Linked List is:" ) printList(head, ahead) # This code is contributed by rutvik_56 |
Output:
Result Linked List is: Traversal using Next Pointer 5, 10, 2, 3, Traversal using Arbit Pointer 2, 3, 5, 10,
Time Complexity: O(n log n), where n is the number of nodes in the Linked list.
Space Complexity: O(1). We are not using any extra space.
Please refer complete article on Point to next higher value node in a linked list with an arbitrary pointer for more details!
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