Remove all even parity nodes from a circular doubly linked list
In this problem, we are dealing with a circular doubly linked list that contains integer values. The task is to remove nodes from the circular doubly linked list that have an even parity of set bits in their binary representation.
Problem Statement and Description
The problem requires us to remove nodes from the circular doubly linked list based on a specific condition. We need to identify the nodes whose binary representation has an even number of set bits (bits with a value of 1) and remove them while maintaining the circular doubly linked structure of the list.
Example
Let's understand the problem with an example. Consider the following circular doubly linked list:
Original Circular Doubly Linked List:
From Front to Tail: 3 > 12 > 32 > 14 > 27 > 19 > 1 > 18
From Tail to Front: 18 > 1 > 19 > 27 > 14 > 32 > 12 > 3
After removing nodes with even parity of set bits, the circular doubly linked list should look like this:
Modified Circular Doubly Linked List:
From Front to Tail: 32 > 14 > 19 > 1
From Tail to Front: 1 > 19 > 14 > 32
Idea to Solve
To solve this problem, we can follow these steps:
 Traverse the circular doubly linked list using the
temp
pointer.  For each node, calculate the number of set bits (bits with a value of 1) in its binary representation using the
coutSetBit
function.  Check if the count of set bits is even. If it is, mark the node for removal.
 Traverse the circular doubly linked list again using the
temp
pointer, this time with the objective of removing the marked nodes.  Depending on whether the current node (
auxiliary
) is the front, tail, or an intermediate node, remove the node from the list while maintaining the circular doubly linked structure.
Pseudocode
removeNode():
if front is null:
return
initialize temp as front
initialize auxiliary as front
while temp is not null:
if coutSetBit(temp.data) is even:
auxiliary = temp
temp = temp.next
if auxiliary is front:
// Remove front node
if auxiliary is tail:
// Removing a single node
tail = null
front = null
temp = null
else:
front = temp
front.back = tail
tail.next = temp
else if auxiliary is tail:
// Remove tail node
auxiliary.back.next = front
front.back = auxiliary.back
tail = auxiliary.back
else:
// Remove intermediate node
auxiliary.back.next = auxiliary.next
auxiliary.next.back = auxiliary.back
auxiliary = null
else:
// visit to next node
temp = temp.next
if temp is front:
// Stop the process
temp = null
Algorithm Explanation
 Traverse the circular doubly linked list using the
temp
pointer.  Calculate the number of set bits in the binary representation of the current node's data using the
coutSetBit
function.  Check if the count of set bits is even. If it is, mark the current node for removal.
 Depending on whether the current node (
auxiliary
) is the front, tail, or an intermediate node, remove the node from the list while maintaining the circular doubly linked structure.  If the current node's data doesn't have an even number of set bits, move
temp
to the next node.  If
temp
has reached back to the front, terminate the loop by settingtemp
to null.
Code Solution

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Time Complexity

In the
coutSetBit
function, the number of set bits is counted using bitwise operations. This operation is executed in constant time, regardless of the input value. Thus, the time complexity of thecoutSetBit
function is O(1). 
In the
removeNode
function, we traverse the circular doubly linked list once to identify nodes with even parity of set bits and then again to remove these nodes. Both traversals have a time complexity of O(N), where N is the number of nodes in the linked list.
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