Length of longest palindrome list in a linked list using O(1) extra space
Last Updated :
06 Apr, 2023
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.
C++
// C++ program to find longest palindrome
// sublist in a list in O(1) time.
#include<bits/stdc++.h>
using namespace std;
//structure of the linked list
struct Node
{
int data;
struct Node* next;
};
// function for counting the common elements
int countCommon(Node *a, Node *b)
{
int count = 0;
// loop to count common in the list starting
// from node a and b
for (; a && b; a = a->next, b = b->next)
// increment the count for same values
if (a->data == b->data)
++count;
else
break;
return count;
}
// Returns length of the longest palindrome
// sublist in given list
int maxPalindrome(Node *head)
{
int result = 0;
Node *prev = NULL, *curr = head;
// loop till the end of the linked list
while (curr)
{
// The sublist from head to current
// reversed.
Node *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
Node *newNode(int key)
{
Node *temp = new Node;
temp->data = key;
temp->next = NULL;
return temp;
}
/* Driver program to test above functions*/
int main()
{
/* Let us create a linked lists to test
the functions
Created list is a: 2->4->3->4->2->15 */
Node *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);
cout << maxPalindrome(head) << endl;
return 0;
}
// Java program to find longest palindrome
// sublist in a list in O(1) time.
class GfG
{
//structure of the linked list
static class Node
{
int data;
Node next;
}
// function for counting the common elements
static int countCommon(Node a, Node b)
{
int count = 0;
// loop to count common in the list starting
// from node a and b
for (; a != null && b != null;
a = a.next, b = b.next)
// increment the count for same values
if (a.data == b.data)
++count;
else
break;
return count;
}
// Returns length of the longest palindrome
// sublist in given list
static int maxPalindrome(Node head)
{
int result = 0;
Node prev = null, curr = head;
// loop till the end of the linked list
while (curr != null)
{
// The sublist from head to current
// reversed.
Node 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 = Math.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 = Math.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
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}
/* Driver code*/
public static void main(String[] args)
{
/* Let us create a linked lists to test
the functions
Created list is a: 2->4->3->4->2->15 */
Node 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);
System.out.println(maxPalindrome(head));
}
}
// This code is contributed by
// Prerna Saini.
# 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
// C# program to find longest palindrome
// sublist in a list in O(1) time.
using System;
class GfG
{
//structure of the linked list
public class Node
{
public int data;
public Node next;
}
// function for counting the common elements
static int countCommon(Node a, Node b)
{
int count = 0;
// loop to count common in the list starting
// from node a and b
for (; a != null && b != null;
a = a.next, b = b.next)
// increment the count for same values
if (a.data == b.data)
++count;
else
break;
return count;
}
// Returns length of the longest palindrome
// sublist in given list
static int maxPalindrome(Node head)
{
int result = 0;
Node prev = null, curr = head;
// loop till the end of the linked list
while (curr != null)
{
// The sublist from head to current
// reversed.
Node 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 = Math.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 = Math.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
static Node newNode(int key)
{
Node temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}
/* Driver code*/
public static void Main(String []args)
{
/* Let us create a linked lists to test
the functions
Created list is a: 2->4->3->4->2->15 */
Node 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);
Console.WriteLine(maxPalindrome(head));
}
}
// This code is contributed by Arnab Kundu
<script>
// Javascript program to find longest palindrome
// sublist in a list in O(1) time.
// structure of the linked list
class Node {
constructor() {
this.data = 0;
this.next = null;
}
}
// function for counting the common elements
function countCommon(a, b) {
var count = 0;
// loop to count common in the list starting
// from node a and b
for (; a != null && b != null; a = a.next, b = b.next)
// increment the count for same values
if (a.data == b.data)
++count;
else
break;
return count;
}
// Returns length of the longest palindrome
// sublist in given list
function maxPalindrome(head) {
var result = 0;
var prev = null, curr = head;
// loop till the end of the linked list
while (curr != null) {
// The sublist from head to current
// reversed.
var 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 = Math.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 = Math.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
function newNode(key) {
var temp = new Node();
temp.data = key;
temp.next = null;
return temp;
}
/* Driver code */
/*
Let us create a linked lists to
test the functions Created list is a:
2->4->3->4->2->15
*/
var 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);
document.write(maxPalindrome(head));
// This code contributed by aashish1995
</script>
Time Complexity : O(n2)
Auxiliary Space: O(1), since no extra space is used.
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.
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