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

For each node
node
in the doubly linked list: a. Check ifnode.next
exists and if its data is the same asnode.data
. b. If true, call theisRotationalPalindrome(node)
function to determine if a rotational palindrome starts fromnode
. 
In the
isRotationalPalindrome(front)
function: a. Initializereverse
to the previous node offront
andforward
to the second next node offront
. b. Handle cases wherereverse
orforward
are null by wrapping around to the opposite end of the list. c. Whilereverse
andforward
are not equal andfront
is not equal toreverse
andforward
is not equal tofront
: i. Compare the data offorward
andreverse
. If they are not equal, returnfalse
. ii. Updatereverse
to the previous node or the tail if it's null. iii. Updateforward
to the next node or the head if it's null. d. If the loop terminates, returntrue
, indicating a rotational palindrome. 
In the
showResult(front)
function: a. Initializenode
tofront.next
. If it's null, setnode
tothis.head
. b. Display the elements of the rotational palindrome, handling the circular nature of the list. 
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 theisRotationalPalindrome()
function. c. If a palindrome is found, display it using theshowResult()
function.
Program List

1) Find the circular palindrome in doubly linked list in java
2) Find the circular palindrome in doubly linked list in c++
3) Find the circular palindrome in doubly linked list in c
4) Find the circular palindrome in doubly linked list in c#
5) Find the circular palindrome in doubly linked list in php
6) Find the circular palindrome in doubly linked list in python
7) Find the circular palindrome in doubly linked list in ruby
8) Find the circular palindrome in doubly linked list in scala
9) Find the circular palindrome in doubly linked list in swift
10) Find the circular palindrome in doubly linked list in kotlin
11) Find the circular palindrome in doubly linked list in typescript
12) Find the circular palindrome in doubly linked list in golang
13) Find the circular palindrome in doubly linked list in vb.net
14) Find the circular palindrome in doubly linked list in node js
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).
Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.
New Comment