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Python Program For Pointing Arbit Pointer To Greatest Value Right Side Node In A...

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Python Program For Pointing Arbit Pointer To Greatest Value Right Side Node In A Linked List

  • Last Updated : 18 May, 2022

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.

listwithArbit1

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. The Time Complexity of this solution is O(n2).

An Efficient Solution can work in O(n) time. Below are the steps.  

  1. Reverse the given linked list.
  2. 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.
  3. 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("None" , 
"        ",end = "")
if (node.arbit != None):
print(node.arbit.data, end = "")
else :
print("None", 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

Output: 

Resultant Linked List is: 
Node    Next Pointer    Arbit Pointer
5               10              10
10              2               3
2               3               3
3               NULL            NULL

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

Output: 

Resultant 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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