Delete all the even nodes of a Circular Linked List
In the problem of deleting all even nodes from a circular linked list, we are dealing with a circular linked list data structure. The objective is to remove nodes from the list that contain even values, while maintaining the circular nature of the linked list.
Problem Statement
Given a circular linked list, we need to delete all nodes that contain even values. After the deletion, the circular structure of the linked list should be preserved.
Example
Consider a circular linked list with the following elements: 12 -> 21 -> 30 -> 41 -> 53 -> 60 -> 79 -> 88. After deleting all even nodes, the list becomes: 21 -> 41 -> 53 -> 79.
Idea to Solve the Problem


To delete all even nodes from a circular linked list, we need to iterate through the list and identify nodes with even values. For each even node, we'll perform the deletion while maintaining the circular structure.
Pseudocode
Here's the pseudocode to delete all even nodes from a circular linked list:
function deleteEvenNodes(circularLinkedList):
if circularLinkedList.head is NULL:
return
temp = circularLinkedList.head.next
auxiliary = circularLinkedList.head
hold = NULL
while temp is not circularLinkedList.head:
hold = temp
temp = temp.next
if hold.data is even:
auxiliary.next = temp
hold = NULL
else:
auxiliary = hold
if circularLinkedList.head.data is even:
temp = circularLinkedList.head
if circularLinkedList.head.next is circularLinkedList.head:
circularLinkedList.head = NULL
else:
circularLinkedList.head = circularLinkedList.head.next
auxiliary.next = circularLinkedList.head
temp = NULL
Algorithm Explanation
- Check if the linked list is empty. If it is, return.
- Initialize
temp
,auxiliary
, andhold
pointers for traversal. - Traverse the list using the
temp
pointer. If the current node's data is even, assign it tohold
; otherwise, assign it toauxiliary
. - Move to the next node (
temp = temp.next
). - If the traversal reaches the end of the list, update the
temp
pointer tocircularLinkedList.head
. - If
hold
is notNULL
, perform the deletion by updating theauxiliary.next
totemp
and freeing memory. - If the head node's data is even, perform the deletion by updating the
head
andauxiliary.next
pointers, and freeing memory.
Code Solution
-
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Time Complexity
The time complexity of deleting all even nodes from a circular linked list is O(n), where n is the number of nodes in the circular linked list. This is because we need to traverse the entire list to identify and remove even nodes. The space complexity is O(1) as we are using only a constant amount of extra space for pointers.
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