Length of longest palindrome list in a linked list using O(1) extra space

Given a linked list, find length of the longest palindrome list that exist 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

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution could be to copy linked list content to array and then find longest palindromic subarray in 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 given linked list and one by one reverse every prefix of linked list from left. After reversing a prefix, we find the longest common list beginning from reversed prefix and list after the reversed prefix.

Below is the implementation of 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 coomon 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

// 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 coomon 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

# 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 coomon 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#

// 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 coomon 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 

Output :

Time Complexity : O(n²)

Note that the above code modifies the given linked list and may not work if modifications to linked list are not allowed. However we can finally do one more reverse to get original list back.

This article is contributed by Niteesh kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python

C#

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