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Python Program For Removing Every K-th Node Of The Linked List

  • Last Updated : 18 May, 2022

Given a singly linked list, Your task is to remove every K-th node of the linked list. Assume that K is always less than or equal to length of Linked List.
Examples :

Input: 1->2->3->4->5->6->7->8  
        k = 3
Output: 1->2->4->5->7->8
As 3 is the k-th node after its deletion list 
would be 1->2->4->5->6->7->8
And now 4 is the starting node then from it, 6 
would be the k-th node. So no other kth node 
could be there.So, final list is:
1->2->4->5->7->8.

Input: 1->2->3->4->5->6  
       k = 1
Output: Empty list 
All nodes need to be deleted

The idea is to traverse the list from the beginning and keep track of nodes visited after the last deletion. Whenever count becomes k, delete the current node and reset the count as 0.

Traverse list and do following
   (a) Count node before deletion.
   (b) If (count == k) that means current 
        node is to be deleted.
      (i)  Delete current node i.e. do

          //  assign address of next node of 
          // current node to the previous node
          // of the current node.
          prev->next = ptr->next i.e.

       (ii) Reset count as 0, i.e., do count = 0.
   (c) Update prev node if count != 0 and if
       count is 0 that means that node is a
       starting point.
   (d) Update ptr and continue until all 
       k-th node gets deleted.

Below is the implementation. 

Python3




# Python3 program to delete every 
# k-th Node of a singly linked list.
import math
  
# Linked list Node 
class Node: 
    def __init__(self, data): 
        self.data = data 
        self.next = None
  
# To remove complete list (Needed 
# for case when k is 1)
def freeList(node):
    while (node != None):
        next = node.next
        node = next
    return node
  
# Deletes every k-th node and 
# returns head of modified list.
def deleteKthNode(head, k):
      
    # If linked list is empty
    if (head == None):
        return None
  
    if (k == 1):
        freeList(head)
        return None
      
    # Initialize ptr and prev before 
    # starting traversal.
    ptr = head
    prev = None
  
    # Traverse list and delete every 
    # k-th node
    count = 0
    while (ptr != None):
          
        # Increment Node count
        count = count + 1
  
        # Check if count is equal to k
        # if yes, then delete current Node
        if (k == count):
              
            # Put the next of current Node in
            # the next of previous Node
            # delete(prev.next)
            prev.next = ptr.next
  
            # Set count = 0 to reach further
            # k-th Node
            count = 0
          
        # Update prev if count is not 0
        if (count != 0):
            prev = ptr
  
        ptr = prev.next
      
    return head
  
# Function to print linked list 
def displayList(head):
    temp = head
    while (temp != None):
        print(temp.data, 
              end = ' ')
        temp = temp.next
      
# Utility function to create 
# a new node.
def newNode( x):
    temp = Node(x)
    temp.data = x
    temp.next = None
    return temp
  
# Driver Code
if __name__=='__main__'
      
    # Start with the empty list 
    head = newNode(1)
    head.next = newNode(2)
    head.next.next = newNode(3)
    head.next.next.next = newNode(4)
    head.next.next.next.next = 
    newNode(5)
    head.next.next.next.next.next = 
    newNode(6)
    head.next.next.next.next.next.next = 
    newNode(7)
    head.next.next.next.next.next.next.next = 
    newNode(8)
  
    k = 3
    head = deleteKthNode(head, k)
  
    displayList(head)
# This code is contributed by Srathore

Output:  

1 2 4 5 7 8

Time Complexity: O(n)

Please refer complete article on Remove every k-th node of the linked list for more details!


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