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Python Program For Finding The Length Of Longest Palindrome List In A Linked List Using O(1) Extra Space

  • Last Updated : 18 Apr, 2022

Given a linked list, find the length of the longest palindrome list that exists in that linked list. 
Examples: 
 

Input  : List = 2->3->7->3->2->12->24
Output : 5
The longest palindrome list is 2->3->7->3->2

Input  : List = 12->4->4->3->14
Output : 2
The longest palindrome list is 4->4

 

A simple solution could be to copy linked list content to array and then find the longest palindromic subarray in the array, but this solution is not allowed as it requires extra space.
The idea is based on iterative linked list reverse process. We iterate through the given a linked list and one by one reverse every prefix of the linked list from the left. After reversing a prefix, we find the longest common list beginning from reversed prefix and the list after the reversed prefix. 
Below is the implementation of the above idea.
 

Python




# Python program to find longest palindrome
# sublist in a list in O(1) time.
 
# Linked List node
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
 
# function for counting the common elements
def countCommon(a, b) :
 
    count = 0
 
    # loop to count common in the list starting
    # from node a and b
    while ( a != None and b != None ) :
 
        # increment the count for same values
        if (a.data == b.data) :
            count = count + 1
        else:
            break
         
        a = a.next
        b = b.next
 
    return count
 
# Returns length of the longest palindrome
# sublist in given list
def maxPalindrome(head) :
 
    result = 0
    prev = None
    curr = head
 
    # loop till the end of the linked list
    while (curr != None) :
     
        # The sublist from head to current
        # reversed.
        next = curr.next
        curr.next = prev
 
        # check for odd length
        # palindrome by finding
        # longest common list elements
        # beginning from prev and
        # from next (We exclude curr)
        result = max(result,
                    2 * countCommon(prev, next) + 1)
 
        # check for even length palindrome
        # by finding longest common list elements
        # beginning from curr and from next
        result = max(result,
                    2 * countCommon(curr, next))
 
        # update prev and curr for next iteration
        prev = curr
        curr = next
     
    return result
 
# Utility function to create a new list node
def newNode(key) :
 
    temp = Node(0)
    temp.data = key
    temp.next = None
    return temp
 
# Driver code
 
# Let us create a linked lists to test
# the functions
# Created list is a: 2->4->3->4->2->15
head = newNode(2)
head.next = newNode(4)
head.next.next = newNode(3)
head.next.next.next = newNode(4)
head.next.next.next.next = newNode(2)
head.next.next.next.next.next = newNode(15)
 
print(maxPalindrome(head))
 
# This code is contributed by Arnab Kundu

Output : 

5

Time Complexity : O(n2)
Note that the above code modifies the given linked list and may not work if modifications to the linked list are not allowed. However, we can finally do one more reverse to get an original list back. Please refer complete article on Length of longest palindrome list in a linked list using O(1) extra space for more details!


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