C Program To 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("");
}

// 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);
        }

        // Link the previous list with
        // remaining nodes
        curr->next = t;

        // Set current pointer for next
        // iteration
        curr = t;
    }
}

// Driver code
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
            Given Linked list is :",
            M, N);
    printList(head);

    skipMdeleteN(head, M, N);

    printf(
    "Linked list after deletion is :");
    printList(head);

    return 0;
}

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!