Remove every kth node of the linked list
The problem aims to remove every kth node from a singly linked list. In this scenario, we are provided with a linked list, and our goal is to delete every kth node, where k is a positive integer. The problem involves traversing the linked list and removing nodes at specific positions.
Problem Description
Given a singly linked list and a positive integer k, we need to modify the linked list by removing every kth node from it. The positions of the nodes to be removed are determined by the value of k, and the removal process should maintain the order of the remaining nodes.
Example
Consider the following linked list:
1 > 2 > 3 > 4 > 5 > 6 > 7
If we choose k = 2, the modified linked list after removing every 2nd node will be:
1 > 3 > 5 > 7
Idea to Solve the Problem
To solve this problem, we can traverse the linked list while keeping track of the current node's position. At every kth position, we remove the current node and adjust the links to skip it. This process continues until we reach the end of the linked list.
Pseudocode
removeEveryKthNode(linkedList, k):
if linkedList.head is NULL:
return
current = linkedList.head
previous = NULL
position = 1
while current is not NULL:
if position % k == 0:
if previous is NULL:
linkedList.head = current.next
else:
previous.next = current.next
current = current.next
else:
previous = current
current = current.next
position = position + 1
Algorithm Explanation
 Start with the
current
pointer pointing to the head of the linked list, and initializeprevious
as NULL.  Initialize a
position
variable to keep track of the current position.  Traverse through the linked list using the
current
pointer: a. Check if the current position is a multiple of k. If yes, it's a kth position, so remove the current node. b. If the position is not a multiple of k, update theprevious
pointer and move thecurrent
pointer to the next node.  After each iteration, increment the
position
.  If the position is a multiple of k, adjust the links to remove the current node.
 Continue these steps until the end of the linked list.
Code Solution

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Time Complexity
The time complexity of this algorithm is O(n), where n is the number of nodes in the linked list. We traverse through each node once to remove the kth nodes.
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