Delete every Kth node from circular linked list
The problem at hand involves deleting every k-th node from a circular linked list, while preserving the circular structure of the list. In a circular linked list, the last node's next pointer points back to the first node, creating a closed loop.
Problem Statement
Given a circular linked list, we need to delete every k-th node from the list while keeping the circular structure intact.
Example
Consider a circular linked list with the following elements: 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8. If k is 2, after deleting every 2nd node, the list becomes: 1 -> 3 -> 5 -> 7.


Idea to Solve the Problem
To delete every k-th node from the circular linked list, we can traverse the list while keeping track of the current position. If the current position is a multiple of k, we delete the node. We'll need to update pointers to maintain the circular structure.
Pseudocode
Here's the pseudocode to delete every k-th node from a circular linked list:
function deleteKthNodes(circularLinkedList, k):
if circularLinkedList.head is NULL:
return
if k <= 0:
return
counter = 0
temp = circularLinkedList.head
hold = NULL
prev = NULL
while temp is not NULL:
counter = counter + 1
if counter % k == 0:
hold = temp
else:
prev = temp
temp = temp.next
if hold is not NULL:
if prev is not NULL:
prev.next = temp
hold = NULL
if temp is circularLinkedList.head:
return
Algorithm Explanation
- Check if the linked list is empty or if k is less than or equal to 0. If any of these conditions are met, return.
- Initialize
counter
to keep track of the current position,temp
to traverse the list, andhold
to store the node to be deleted. - Traverse the list using the
temp
pointer. If the current position is a multiple of k, update thehold
pointer. - If the current position is not a multiple of k, update the
prev
pointer. - Move to the next node (
temp = temp.next
). - If
hold
is not NULL, it means we need to delete a node. Update theprev.next
pointer to skip the node to be deleted. - If
temp
reaches the head of the circular list, return.
Code Solution
-
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Time Complexity
The time complexity of deleting every k-th node from a circular linked list is O(n), where n is the number of nodes in the circular linked list. Since we need to traverse the entire list once, the time complexity is linear. The space complexity is O(1) as we are using only a constant amount of extra space for pointers.
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