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

07 Jul, 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.
C++
Java
Python
C#
Javascript
// 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 liststruct Node{    int data;    struct Node* next;}; // function for counting the common elementsint 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 listint 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 nodeNode *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 liststatic class Node{    int data;    Node next;} // function for counting the common elementsstatic 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 liststatic 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 nodestatic 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 nodeclass Node:    def __init__(self, data):        self.data = data        self.next = None # function for counting the common elementsdef 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 listdef 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 nodedef 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->15head = 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 listpublic class Node{    public int data;    public Node next;} // function for counting the common elementsstatic 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 liststatic 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 nodestatic 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>
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.
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