Python Program For Pointing Arbit Pointer To Greatest Value Right Side Node In A Linked List
Last Updated :
23 Jul, 2025
Given singly linked list with every node having an additional “arbitrary” pointer that currently points to NULL. We need to make the “arbitrary” pointer to the greatest value node in a linked list on its right side.
A Simple Solution is to traverse all nodes one by one. For every node, find the node which has the greatest value on the right side and change the next pointer.
Below is the implementation of the above approach:
Python3
# Python Program to point arbit pointers
# to highest value on its right
# Node class
class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
self.arbit = None
# Function to reverse the linked list
def reverse(head):
prev = None
current = head
next = None
while (current != None):
next = current.next
current.next = prev
prev = current
current = next
return prev
# This function populates arbit pointer
# in every node to the greatest value
# to its right.
def populateArbit(head):
# Reverse given linked list
head = reverse(head)
# Initialize pointer to maximum value node
max = head
# Traverse the reversed list
temp = head.next
while (temp != None):
# Connect max through arbit pointer
temp.arbit = max
# Update max if required
if (max.data < temp.data):
max = temp
# Move ahead in reversed list
temp = temp.next
# Reverse modified linked list and return head.
return reverse(head)
# Utility function to print result linked list
def printNextArbitPointers(node):
print("Node ", "Next Pointer " , "Arbit Pointer")
while (node != None):
print(node.data , " ", end = "")
if (node.next != None):
print(node.next.data , " ", end = "")
else :
print("NULL" , " ",end = "")
if (node.arbit != None):
print(node.arbit.data, end = "")
else :
print("NULL", end = "")
print("")
node = node.next
# Function to create a new node with given data
def newNode(data):
new_node = Node(0)
new_node.data = data
new_node.next = None
return new_node
# Driver code
head = newNode(5)
head.next = newNode(10)
head.next.next = newNode(2)
head.next.next.next = newNode(3)
head = populateArbit(head)
print("Resultant Linked List is: ")
printNextArbitPointers(head)
# This code is contributed by Susobhan Akhuli
OutputResultant Linked List is:
Node Next Pointer Arbit Pointer
5 10 10
10 2 3
2 3 3
3 NULL NULL
The Time Complexity of this above solution is O(n2).
Space complexity:
The space complexity of the program is O(1) as it uses a constant amount of extra space irrespective of the input size.
An Efficient Solution can work in O(n) time. Below are the steps.
- Reverse the given linked list.
- Start traversing the linked list and store the maximum value node encountered so far. Make arbit of every node to point to max. If the data in the current node is more than the max node so far, update max.
- Reverse modified linked list and return head.
Following is the implementation of the above steps.
Python
# Python Program to point arbit pointers
# to highest value on its right
# Node class
class Node:
# Constructor to initialize the
# node object
def __init__(self, data):
self.data = data
self.next = None
self.arbit = None
# Function to reverse the linked list
def reverse(head):
prev = None
current = head
next = None
while (current != None):
next = current.next
current.next = prev
prev = current
current = next
return prev
# This function populates arbit pointer
# in every node to the greatest value
# to its right.
def populateArbit(head):
# Reverse given linked list
head = reverse(head)
# Initialize pointer to maximum
# value node
max = head
# Traverse the reversed list
temp = head.next
while (temp != None):
# Connect max through arbit
# pointer
temp.arbit = max
# Update max if required
if (max.data < temp.data):
max = temp
# Move ahead in reversed list
temp = temp.next
# Reverse modified linked list and
# return head.
return reverse(head)
# Utility function to print result
# linked list
def printNextArbitPointers(node):
print("Node ",
"Next Pointer " ,
"Arbit Pointer")
while (node != None):
print(node.data ,
" ",
end = "")
if (node.next != None):
print(node.next.data ,
" ", end = "")
else :
print("NULL" ,
" ",end = "")
if (node.arbit != None):
print(node.arbit.data, end = "")
else :
print("NULL", end = "")
print("")
node = node.next
# Function to create a new node
# with given data
def newNode(data):
new_node = Node(0)
new_node.data = data
new_node.next = None
return new_node
# Driver code
head = newNode(5)
head.next = newNode(10)
head.next.next = newNode(2)
head.next.next.next = newNode(3)
head = populateArbit(head)
print("Resultant Linked List is: ")
printNextArbitPointers(head)
# This code is contributed by Arnab Kundu
# This code is modified by Susobhan Akhuli
OutputResultant Linked List is:
Node Next Pointer Arbit Pointer
5 10 10
10 2 3
2 3 3
3 NULL NULL
The time complexity of this program is O(n), where n is the number of nodes in the linked list.
The space complexity of this program is also O(n), as it creates a new node for each element in the list and stores them in memory.
Recursive Solution:
We can recursively reach the last node and traverse the linked list from the end. The recursive solution doesn’t require reversing of the linked list. We can also use a stack in place of recursion to temporarily hold nodes. Thanks to Santosh Kumar Mishra for providing this solution.
Python3
# Python3 program to point arbit pointers
# to highest value on its right
''' Link list node '''
# Node class
class newNode:
# Constructor to initialize the
# node object
def __init__(self, data):
self.data = data
self.next = None
self.arbit = None
# This function populates arbit pointer
# in every node to the greatest value
# to its right.
maxNode = newNode(None)
def populateArbit(head):
# using static maxNode to keep track
# of maximum orbit node address on
# right side
global maxNode
# if head is null simply return the list
if (head == None):
return
''' if head.next is null it means we
reached at the last node just update
the max and maxNode '''
if (head.next == None):
maxNode = head
return
''' Calling the populateArbit to the
next node '''
populateArbit(head.next)
''' updating the arbit node of the
current node with the maximum
value on the right side '''
head.arbit = maxNode
''' if current Node value id greater
then the previous right node then
update it '''
if (head.data > maxNode.data and
maxNode.data != None ):
maxNode = head
return
# Utility function to prresult
# linked list
def printNextArbitPointers(node):
print("Node " ,
"Next Pointer " ,
"Arbit Pointer")
while (node != None):
print(node.data,
" ",
end = "")
if(node.next):
print(node.next.data,
" ",
end = "")
else:
print("NULL", " ",
end = "")
if(node.arbit):
print(node.arbit.data, end = "")
else:
print("NULL", end = "")
print()
node = node.next
# Driver code
head = newNode(5)
head.next = newNode(10)
head.next.next = newNode(2)
head.next.next.next = newNode(3)
populateArbit(head)
print("Resultant Linked List is:")
printNextArbitPointers(head)
# This code is contributed by SHUBHAMSINGH10
OutputResultant Linked List is:
Node Next Pointer Arbit Pointer
5 10 10
10 2 3
2 3 3
3 NULL NULL
Please refer complete article on Point arbit pointer to greatest value right side node in a linked list for more details!
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