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Code Doubly linked list

Detect rotational palindrome in doubly linked list

The problem at hand involves detecting rotational palindromes in a doubly linked list. A rotational palindrome is a sequence of elements that remains unchanged when the list is rotated either to the left or right. In other words, if we take a subsection of the linked list and rotate it, the elements in that subsection should still read the same forwards and backwards.

Problem Statement

Given a doubly linked list, we need to determine whether there exists a rotational palindrome within the list and if so, identify and display the elements of that palindrome.

Example

Consider the following linked list:

2 <-> 3 <-> 1 <-> 1 <-> 1 <-> 1 <-> 3 <-> 2 <-> 1

Rotating this list to the right or left by some positions results in:

1 <-> 1 <-> 3 <-> 2 <-> 1 <-> 2 <-> 3 <-> 1 <-> 1

As you can see, the rotated sequence "1 1 3 2 1 2 3 1 1" is a palindrome.

Idea to Solve

The idea to solve this problem involves checking each node in the doubly linked list as a potential starting point for a rotational palindrome. For each starting point, we'll examine its neighboring elements and determine if they form a rotational palindrome.

Algorithm

  1. For each node node in the doubly linked list: a. Check if node.next exists and if its data is the same as node.data. b. If true, call the isRotationalPalindrome(node) function to determine if a rotational palindrome starts from node.

  2. In the isRotationalPalindrome(front) function: a. Initialize reverse to the previous node of front and forward to the second next node of front. b. Handle cases where reverse or forward are null by wrapping around to the opposite end of the list. c. While reverse and forward are not equal and front is not equal to reverse and forward is not equal to front: i. Compare the data of forward and reverse. If they are not equal, return false. ii. Update reverse to the previous node or the tail if it's null. iii. Update forward to the next node or the head if it's null. d. If the loop terminates, return true, indicating a rotational palindrome.

  3. In the showResult(front) function: a. Initialize node to front.next. If it's null, set node to this.head. b. Display the elements of the rotational palindrome, handling the circular nature of the list.

  4. In the palindrome() function: a. Handle cases where the linked list is empty or has only one element. b. Iterate through the linked list and check for potential rotational palindromes using the isRotationalPalindrome() function. c. If a palindrome is found, display it using the showResult() function.

Program List

Time Complexity

The time complexity of this algorithm mainly depends on the isRotationalPalindrome() function, which iterates through the linked list to check for rotational palindromes. In the worst case, it would iterate through the entire list twice, resulting in O(n) time complexity, where n is the number of elements in the linked list. The overall time complexity of the entire algorithm remains O(n).

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