Remove all even parity nodes from a circular single linked list
In this problem, we are dealing with a circular singly linked list that contains integer values. The task is to remove nodes from the circular 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 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 structure of the linked list.
Example
Let's understand the problem with an example. Consider the following circular linked list:
Original Circular Linked List: 3 > 12 > 32 > 14 > 27 > 19 > 1 > 18 > 3 (back to the start)
After removing nodes with even parity of set bits, the circular linked list should look like this:
Modified Circular Linked List: 32 > 14 > 19 > 1 > 32 (back to the start)
Idea to Solve
To solve this problem, we can follow these steps:
 Traverse the circular 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 linked list again using the
temp
pointer, this time with the objective of removing the marked nodes.  For each marked node, remove it from the list while maintaining the circular structure.
Pseudocode
removeNode():
if head is null:
return
initialize temp as head
initialize auxiliary as head
initialize point as null
while temp is not null:
if coutSetBit(temp.data) is even:
auxiliary = temp
temp = temp.next
if auxiliary is head:
// Remove head node
if auxiliary is tail:
// Removing a single node
tail = null
head = null
temp = null
else:
head = temp
tail.next = temp
else if auxiliary is tail:
if point is not null:
// Remove last node
point.next = head
tail = point
else:
if point is not null:
// Remove intermediate node
point.next = temp
auxiliary = null
else:
point = temp
temp = temp.next
if temp is head:
// Stop the process
temp = null
Algorithm Explanation
 Traverse the circular 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 head, tail, or an intermediate node, remove the node from the list while maintaining the circular structure.  If the current node's data doesn't have an even number of set bits, update the
point
pointer totemp
and movetemp
to the next node.  If
temp
has reached back to the head, 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 linked list twice. The first traversal identifies nodes with even parity of set bits. The second traversal removes these nodes while maintaining the circular structure. Both traversals have a time complexity of O(N), where N is the number of nodes in the linked list.
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