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Python Program To Delete N Nodes After M Nodes Of A Linked List

  • Last Updated : 18 May, 2022

Given a linked list and two integers M and N. Traverse the linked list such that you retain M nodes then delete next N nodes, continue the same till end of the linked list.
Difficulty Level: Rookie 
Examples:

Input:
M = 2, N = 2
Linked List: 1->2->3->4->5->6->7->8
Output:
Linked List: 1->2->5->6

Input:
M = 3, N = 2
Linked List: 1->2->3->4->5->6->7->8->9->10
Output:
Linked List: 1->2->3->6->7->8

Input:
M = 1, N = 1
Linked List: 1->2->3->4->5->6->7->8->9->10
Output:
Linked List: 1->3->5->7->9

The main part of the problem is to maintain proper links between nodes, make sure that all corner cases are handled. Following is C implementation of function skipMdeleteN() that skips M nodes and delete N nodes till end of list. It is assumed that M cannot be 0.

Python




# Python program to delete M nodes 
# after N nodes
# Node class 
class Node:
  
    # Constructor to initialize 
    # the node object
    def __init__(self, data):
        self.data = data
        self.next = None
  
class LinkedList:
  
    # Function to initialize head
    def __init__(self):
        self.head = None
  
    # Function to insert a new node 
    # at the beginning
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
  
    # Utility function to print the 
    # linked LinkedList
    def printList(self):
        temp = self.head
        while(temp):
            print temp.data,
            temp = temp.next
  
    def skipMdeleteN(self, M, N):
        curr = self.head
          
        # The main loop that traverses 
        # through the whole list
        while(curr):
            # Skip M nodes
            for count in range(1, M):
                if curr is None:
                    return 
                curr = curr.next
                      
            if curr is None :
                return 
  
            # Start from next node and delete 
            # N nodes
            t = curr.next 
            for count in range(1, N+1):
                if t is None:
                    break
                t = t.next
      
            # Link the previous list with 
            # remaining nodes
            curr.next = t
  
            # Set Current pointer for next 
            # iteration
            curr =
  
# Driver code
  
# Create following linked list
# 1->2->3->4->5->6->7->8->9->10
llist = LinkedList()
M = 2 
N = 3
llist.push(10)
llist.push(9)
llist.push(8)
llist.push(7)
llist.push(6)
llist.push(5)
llist.push(4)
llist.push(3)
llist.push(2)
llist.push(1)
  
print "M = %d, N = %d
Given Linked List is:" %(M, N)
llist.printList()
print 
  
llist.skipMdeleteN(M, N)
  
print "Linked list after deletion is"
llist.printList()
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output: 

M = 2, N = 3
Given Linked list is :
1 2 3 4 5 6 7 8 9 10
Linked list after deletion is :
1 2 6 7

Time Complexity:
O(n) where n is number of nodes in linked list.

Auxiliary Space: O(1)

Please refer complete article on Delete N nodes after M nodes of a linked list for more details!


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