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

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  • Last Updated : 11 Jan, 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++




// 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;
}


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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