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:

1. Traverse the circular doubly linked list using the temp pointer.
2. For each node, calculate the number of set bits (bits with a value of 1) in its binary representation using the coutSetBit function.
3. Check if the count of set bits is even. If it is, mark the node for removal.
4. Traverse the circular doubly linked list again using the temp pointer, this time with the objective of removing the marked nodes.
5. 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

1. Traverse the circular doubly linked list using the temp pointer.
2. Calculate the number of set bits in the binary representation of the current node's data using the coutSetBit function.
3. Check if the count of set bits is even. If it is, mark the current node for removal.
4. 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.
5. If the current node's data doesn't have an even number of set bits, move temp to the next node.
6. If temp has reached back to the front, terminate the loop by setting temp to null.

Code Solution

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Time Complexity

1. 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 the coutSetBit function is O(1).

2. 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.