Javascript 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.

Javascript

<script>
// Javascript program to delete N nodes
// after M nodes of a linked list

// A linked list node
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
}

// Function to insert a node at
// the beginning
function push(head_ref, new_data)
{
    // Allocate node
    var new_node = new 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;

    return head_ref;
}

// Function to print linked list
function printList(head)
{
    var temp = head;
    while (temp != null)
    {
        document.write(temp.data + " ");
        temp = temp.next;
    }
    document.write("<br/>");
}

// Function to skip M nodes and then
// delete N nodes of the linked list.
function skipMdeleteN(head, M , N)
{
    var curr = head, t;
    var count;

    // The main loop that traverses
    // through the whole list
    while (curr != null)
    {
        // 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++)
        {
            var temp = t;
            t.next;
        }

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

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

// Driver code
/* Create following linked list
   1.2.3.4.5.6.7.8.9.10 */
var head = null;
var M = 2, N = 3;
head = push(head, 10);
head = push(head, 9);
head = push(head, 8);
head = push(head, 7);
head = push(head, 6);
head = push(head, 5);
head = push(head, 4);
head = push(head, 3);
head = push(head, 2);
head = push(head, 1);

document.write(
"M = "+M+", N = " + N +
"<br/>Given Linked list is :<br/>");
printList(head);

skipMdeleteN(head, M, N);

document.write(
"<br/>Linked list after deletion is :<br/>");
printList(head);
// This code is contributed by gauravrajput1
</script>

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!