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():
    if head is null or head.next is null:
        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():
    if head is null:
        return
    Initialize temp = head, back = null
    Update tail as head

    Traverse the list and adjust links to reverse nodes

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

Program solution

1) Anticlockwise spiral conversion of doubly linked list in java 2) Anticlockwise spiral conversion of doubly linked list in c++ 3) Anticlockwise spiral conversion of doubly linked list in c 4) Anticlockwise spiral conversion of doubly linked list in c# 5) Anticlockwise spiral conversion of doubly linked list in php 6) Anticlockwise spiral conversion of doubly linked list in python 7) Anticlockwise spiral conversion of doubly linked list in ruby 8) Anticlockwise spiral conversion of doubly linked list in scala 9) Anticlockwise spiral conversion of doubly linked list in swift 10) Anticlockwise spiral conversion of doubly linked list in kotlin 11) Anticlockwise spiral conversion of doubly linked list in vb.net 12) Anticlockwise spiral conversion of doubly linked list in node js 13) Anticlockwise spiral conversion of doubly linked list in golang 14) Anticlockwise spiral conversion of doubly linked list in typescript

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).