Posted on by Kalkicode

# Convert a doubly linked list anticlockwise spiral form

This article we find the solution of transforming a doubly linked list into an anticlockwise spiral form. We will delve into the problem's context, provide a comprehensive explanation using an illustrative example, outline the algorithmic approach, present standard pseudocode, detail the algorithm's implementation, explain the output, and analyze the time complexity. The discussion will be presented in a simple and accessible manner.

## Problem Statement

Given a doubly linked list, the task is to rearrange its elements into an anticlockwise spiral pattern while retaining the linked list's original structure. The challenge lies in manipulating the links between nodes to achieve the desired arrangement.

## Illustrative Example

Consider the following doubly linked list:

``1 <-> 2 <-> 3 <-> 4 <-> 5 <-> 6 <-> 7 <-> 8 <-> 9``

After converting the list into the anticlockwise spiral form, it will appear as:

``9 <-> 1 <-> 8 <-> 2 <-> 7 <-> 3 <-> 6 <-> 4 <-> 5``

This transformation involves reordering the nodes to form an anticlockwise spiral pattern while keeping the links intact.

this is actual transformation.

``````Before anticlockwise spiral conversion
Linked List Head to Tail :  1  2  3  4  5  6  7  8  9
Linked List Tail to Head :  9  8  7  6  5  4  3  2  1
After anticlockwise spiral conversion
Linked List Head to Tail :  9  1  8  2  7  3  6  4  5
Linked List Tail to Head :  5  4  6  3  7  2  8  1  9``````

## Algorithmic Approach

1. Initialize pointers `current` and `last` to the head and tail of the linked list, respectively.
2. Set up auxiliary pointers to facilitate link swapping.
3. Begin the spiral transformation:
• Swap links between `current` and `last` to create the anticlockwise spiral pattern.
• Update pointers and auxiliary pointers as needed.
4. Reverse the entire linked list to finalize the transformation.

## Pseudocode

``````spiralTransform():
return
Initialize last = tail, current = head, auxiliary = null
Set up temp and back pointers

Set head as tail to start the new configuration

Loop until current and last pointers meet or cross:
Handle edge cases when current and last are adjacent
Swap links to form the anticlockwise spiral pattern
Update pointers and auxiliary pointers

Reverse the entire linked list using reverseNode()

reverseNode():
return
Initialize temp = head, back = null

Main():
Initialize a doubly linked list (dll)
Insert nodes into the linked list

Display the list before the transformation
Convert the list to anticlockwise spiral form
Display the transformed list``````

## Time Complexity Analysis

• The `spiralTransform` function iterates through the linked list once, performing constant-time operations per iteration. Thus, its time complexity is O(n), where n is the number of nodes in the list.
• The `reverseNode` function also traverses the entire linked list once, resulting in a time complexity of O(n).

## Comment

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.

Categories
Relative Post