Remove all fibonacci nodes from a circular single linked list

In this problem, we are working with a circular singly linked list containing integer values. The task is to remove all nodes from the circular linked list whose data values are Fibonacci numbers.

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 data values are Fibonacci numbers 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 -> 1 -> 14 -> 0 -> 27 -> 13 -> 89 -> 3 (back to the start)

After removing the Fibonacci nodes, the circular linked list should look like this:

Modified Circular Linked List: 12 -> 32 -> 14 -> 27 -> 12 (back to the start)

Idea to Solve

To solve this problem, we can follow these steps:

1. Traverse the circular linked list to find the maximum value present in the nodes. This value will help us determine the range of Fibonacci numbers we need to generate.
2. Generate a HashSet containing Fibonacci numbers up to the maximum value found in step 1. This HashSet will allow us to quickly check if a node's data is a Fibonacci number.
3. Traverse the circular linked list again and for each node, check if its data value is in the HashSet of Fibonacci numbers. If it is, remove the node from the list while maintaining the circular structure.

Pseudocode

removeFibonacciNode():
    if head is null:
        return
    initialize record as an empty HashSet
    initialize temp as head.next
    initialize max as head.data
    
    // Find the maximum node value
    while temp is not null and temp is not head:
        if max < temp.data:
            max = temp.data
        temp = temp.next
        
    // Generate Fibonacci numbers up to max
    fibNumber(record, max)
    
    initialize auxiliary as head
    initialize point as null
    initialize temp as head
    
    while temp is not null:
        if data of temp is in record:
            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
                    point = null
                else:
                    head = temp
                    tail.next = temp
            else if auxiliary is tail:
                // Remove last node
                if point is not null:
                    point.next = head
                tail = point
            else:
                // Remove intermediate node
                if point is not null:
                    point.next = temp
            auxiliary = null
        else:
            point = temp
            temp = temp.next
            if temp is head:
                temp = null

Algorithm Explanation

1. Traverse the circular linked list to find the maximum value among the nodes. This value will be used to generate Fibonacci numbers.
2. Generate a HashSet called record containing Fibonacci numbers up to the maximum value using the fibNumber function.
3. Traverse the circular linked list again using the temp pointer.
4. Check if the data value of the current node (temp.data) is present in the record HashSet.
5. If it is in the record, store the reference to the current node (temp) in the auxiliary pointer and move temp to the next node.
6. Depending on whether auxiliary is the head, tail, or an intermediate node, remove the node from the list while maintaining the circular structure.
7. If the current node's data is not a Fibonacci number, update the point pointer to temp and move temp to the next node.
8. If temp has reached back to the head, terminate the loop by setting temp to null.

Code Solution

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Time Complexity

1. In the fibNumber function, we generate Fibonacci numbers up to a maximum value. The number of Fibonacci numbers generated depends on the maximum value. Let's denote this maximum value as N. Generating Fibonacci numbers has a time complexity of O(N), and inserting them into the HashSet also has a time complexity of O(N). Thus, the overall time complexity of the fibNumber function is O(N).

2. In the removeFibonacciNode function, we traverse the circular linked list twice. The first traversal finds the maximum value among the nodes, which takes O(N) time. The second traversal removes nodes based on Fibonacci numbers, which also takes O(N) time due to the while loop. The operations within the loop, such as updating pointers, are constant-time operations. Therefore, the overall time complexity of the removeFibonacciNode function is O(N).