Delete N nodes after M nodes of a linked list

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.

C

// C program to delete N nodes after M nodes of a linked list
#include <stdio.h>
#include <stdlib.h>
  
// A linked list node
struct Node
{
    int data;
    struct Node *next;
};
  
/* Function to insert a node at the beginning */
void push(struct Node ** head_ref, int new_data)
{
    /* allocate node */
    struct Node* new_node = (struct Node*) malloc(sizeof(struct Node));
  
    /* put in the data  */
    new_node->data  = new_data;
  
    /* link the old list off the new node */
    new_node->next = (*head_ref);
  
    /* move the head to point to the new node */
    (*head_ref)  = new_node;
}
  
/* Function to print linked list */
void printList(struct Node *head)
{
    struct Node *temp = head;
    while (temp != NULL)
    {
        printf("%d ", temp->data);
        temp = temp->next;
    }
    printf("\n");
}
  
// Function to skip M nodes and then delete N nodes of the linked list.
void skipMdeleteN(struct Node  *head, int M, int N)
{
    struct Node *curr = head, *t;
    int count;
  
    // The main loop that traverses through the whole list
    while (curr)
    {
        // Skip M nodes
        for (count = 1; count<M && curr!= NULL; count++)
            curr = curr->next;
  
        // If we reached end of list, then return
        if (curr == NULL)
            return;
  
        // Start from next node and delete N nodes
        t = curr->next;
        for (count = 1; count<=N && t!= NULL; count++)
        {
            struct Node *temp = t;
            t = t->next;
            free(temp);
        }
        curr->next = t; // Link the previous list with remaining nodes
  
        // Set current pointer for next iteration
        curr = t;
    }
}
  
// Driver program to test above functions
int main()
{
    /* Create following linked list
      1->2->3->4->5->6->7->8->9->10 */
    struct Node* head = NULL;
    int M=2, N=3;
    push(&head, 10);
    push(&head, 9);
    push(&head, 8);
    push(&head, 7);
    push(&head, 6);
    push(&head, 5);
    push(&head, 4);
    push(&head, 3);
    push(&head, 2);
    push(&head, 1);
  
    printf("M = %d, N = %d \nGiven Linked list is :\n", M, N);
    printList(head);
  
    skipMdeleteN(head, M, N);
  
    printf("\nLinked list after deletion is :\n");
    printList(head);
  
    return 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 prit 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 reamining nodes
            curr.next = t
            # Set Current pointer for next iteration
            curr = t 
  
# Driver program to test above function
  
# 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\nGiven Linked List is:" %(M, N)
llist.printList()
print 
  
llist.skipMdeleteN(M, N)
  
print "\nLinked 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.

This article is contributed by Chandra Prakash. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



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Improved By : AbhishekValsan