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Python Program For Pointing To Next Higher Value Node In A Linked List With An Arbitrary Pointer

Last Updated : 20 Mar, 2023
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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 program to populate arbit pointers
# to next higher value using merge sort
head = None
# Link l node
class Node:
    def __init__(self, data):       = data = None
        self.arbit = None
# Utility function to print result
# linked l
def printList(node, anode):
    print("Traversal using Next Pointer")
    while (node != None):
        print(, end = ", ")
        node =
    print("Traversal using Arbit Pointer");
    while (anode != None):
        print(, 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 =
    # 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 ( <=
        result = a
        result.arbit = SortedMerge(a.arbit, b)
        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) = Node(10) = Node(2) = 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


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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