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
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 :
5
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.
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