Good evening! Here's our prompt for today.

Removing the Nth Node From a Linked List's End: A Single-Pass Approach

Get ready to sail through a classic problem: Removing the nth node from the end of a linked list in a single pass. By the end of this lesson, you'll be able to wave goodbye to that pesky nth node and still maintain the list's structure.

The Task

Imagine a chain of dominoes, each representing a node in our linked list. In this scenario, a list might look like this:

1 -> 2 -> 3 -> 4 -> 5

Your mission is to remove the nth domino from the end. So, if we call removeFromEnd(head, 2), the chain would lose the 4 domino, and the head would still be 1.

Ground Rules

Before you jump in, let's set some ground rules:

• Empty List Handling: The linked list could be empty, akin to a chain with no dominoes.
• Nth Node Edge Case: If n exceeds the length of the list, then the chain remains intact. No dominoes are toppled!
• Efficiency Requirements: Aim for a time complexity of O(n) and a space complexity of O(1).

Try to solve this here or in Interactive Mode.

How do I practice this challenge?

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    'PASSED: Removing the 2nd element from the end should get us `1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 8`'

var assert = require('assert');

​

function removeFromEnd(head, idxFromEnd) {

  // remove idxFromEnd-th node

  return head;

}

​

function Node(val) {

  this.val = val;

  this.next = null;

}

​

function LinkedListNode(val) {

  this.val = val;

  this.next = null;

}

​

var list1 = new LinkedListNode(3);

var nodes1 = [4, 5, 6, 7, 8, 9, 10];

createNodes(list1, nodes1);

​

var list2 = new LinkedListNode(1);

var nodes2 = [2, 3, 4, 5, 6, 7, 8];

createNodes(list2, nodes2);

​

function createNodes(head, nodes) {

  for (let i = 0; i < nodes.length; i++) {

    var newNode = new LinkedListNode(nodes[i]);

    head.next = newNode;

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Here's how we would solve this problem...

How do I use this guide?