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# Convert a doubly linked list spiral form

The problem at hand involves converting a given doubly linked list into a spiral form. A doubly linked list is a data structure in which each node contains a data element and two pointers, one pointing to the next node and another pointing to the previous node. The objective is to rearrange the nodes of the doubly linked list in a spiral fashion, combining the first and last nodes, then the second and second-to-last nodes, and so on.

## Problem Statement

Given a doubly linked list, the task is to rearrange its nodes such that they form a spiral pattern. This involves combining the first and last nodes, then the second and second-to-last nodes, and so on, until the middle is reached.

## Example

Consider the following doubly linked list:

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

After converting the doubly linked list into a spiral form, it will look like:

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

## Idea to Solve

The idea to solve this problem is to iteratively manipulate the pointers of the doubly linked list nodes to achieve the desired spiral pattern. This involves combining nodes from the beginning and end of the list, updating their pointers appropriately to form the spiral shape.

## Pseudocode

``````function spiral():
while current is not null and tail is not null:
if current equals tail:
current.next = null
return
else if current.next equals tail:
tail.next = null
return

auxiliary = current.next
current.next = tail
temp = tail.prev
tail.prev = current
tail.next = auxiliary

if auxiliary is not null:
auxiliary.prev = tail

current = auxiliary
tail = temp``````

## Algorithm Explanation

1. Start with two pointers: `current` pointing to the head of the doubly linked list and `tail` pointing to its tail.
2. Loop while both `current` and `tail` are not null:
• If `current` reaches the same node as `tail`, set `current.next` to null and return.
• Else, if `current.next` reaches the same node as `tail`, set `tail.next` to null and return.
• Create an `auxiliary` pointer to store `current.next`.
• Update the pointers to achieve the spiral pattern: `current.next` points to `tail`, `tail.prev` points to `current`, and `tail.next` points to `auxiliary`.
• If `auxiliary` is not null, set its `prev` pointer to `tail`.
• Move `current` to `auxiliary` and update `tail` to its previous node.
3. Repeat this process until the entire spiral pattern is formed.

## Time Complexity

The time complexity of this algorithm is O(n), where n is the number of nodes in the doubly linked list. This is because in the worst case, we iterate through all the nodes once to perform the spiral conversion operation.

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