Why Remove Nth Node from End of List in DSA Typescript?

The Big Idea

What if you could remove the last few items from a list without counting the whole thing first?

The Scenario

Imagine you have a long chain of paper clips linked together. You want to remove the 3rd paper clip from the end, but you only have the chain from the start. You try counting each clip from the start to find the right one to remove.

The Problem

Counting from the start to find the 3rd from the end means you must first know the total number of clips. If the chain is very long, counting twice wastes time and can cause mistakes. You might remove the wrong clip if you miscount.

The Solution

Using the "Remove Nth Node from End of List" method, you can find and remove the target clip in just one pass. This method uses two pointers moving through the chain at different speeds, so you know exactly which clip to remove without counting the whole chain first.

Before vs After

✗ Before

let length = 0;
let current = head;
while (current) {
  length++;
  current = current.next;
}
// Then find (length - n)th node to remove

✓ After

let fast = head;
let slow = head;
for (let i = 0; i < n; i++) fast = fast.next;
while (fast && fast.next) {
  fast = fast.next;
  slow = slow.next;
}
// slow.next is the node to remove

What It Enables

This method lets you quickly and safely remove any node from the end of a list in just one pass, saving time and avoiding errors.

Real Life Example

When managing a playlist of songs, you might want to remove the 2nd last song quickly without counting all songs. This method helps you do that efficiently.

Key Takeaways

Manual counting is slow and error-prone for removing nodes from the end.

Two-pointer technique finds the target node in one pass.

Efficient and safe removal from linked lists saves time and mistakes.

Practice

1. What does the problem Remove Nth Node from End of List ask you to do?

easy

A. Find the value of the nth node from the start

B. Add a new node at the nth position from the start

C. Reverse the linked list up to the nth node

D. Delete the node that is the nth from the end in a linked list

Solution

1. Step 1: Understand the problem statement

   The problem requires removing the node that is nth from the end of a singly linked list.

2. Step 2: Clarify the task

   We do not add or reverse nodes; we specifically remove the nth node counting from the list's end.

3. Final Answer:

   Delete the node that is the nth from the end in a linked list -> Option D

4. Quick Check:

   Remove nth from end = Delete nth from end [OK]

Hint: Focus on 'from end' meaning count backwards [OK]

Common Mistakes:

• Confusing nth from end with nth from start
• Thinking the problem is about reversing the list
• Assuming insertion instead of deletion

2. Which of the following TypeScript function signatures correctly represents a function to remove the nth node from the end of a singly linked list?

easy

A. function removeNthFromEnd(head: ListNode, n: string): ListNode

B. function removeNthFromEnd(head: number[], n: number): number[]

C. function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null

D. function removeNthFromEnd(n: number, head: ListNode): ListNode

Solution

1. Step 1: Check parameter types and order

   The function should take the head of a linked list (ListNode or null) and a number n, returning the possibly updated head.

2. Step 2: Validate correct types and order

   function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null matches this: head is ListNode|null, n is number, return type is ListNode|null.

3. Final Answer:

   function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null -> Option C

4. Quick Check:

   Correct signature uses ListNode and number types [OK]

Hint: Head first, then n; use ListNode type [OK]

Common Mistakes:

• Swapping parameter order
• Using array type instead of ListNode
• Using wrong type for n (string instead of number)

3. What is the output linked list after removing the 2nd node from the end in the list 1 -> 2 -> 3 -> 4 -> 5 -> null?

medium

A. 1 -> 2 -> 4 -> 5 -> null

B. 1 -> 2 -> 3 -> 5 -> null

C. 1 -> 3 -> 4 -> 5 -> null

D. 2 -> 3 -> 4 -> 5 -> null

Solution

1. Step 1: Identify the 2nd node from the end

   Counting from the end: 5 is 1st, 4 is 2nd node from the end.

2. Step 2: Remove the 4th node

   Removing node with value 4 results in list: 1 -> 2 -> 3 -> 5 -> null.

3. Final Answer:

   1 -> 2 -> 3 -> 5 -> null -> Option B

4. Quick Check:

   2nd from end is 4, removed list matches 1 -> 2 -> 3 -> 5 -> null [OK]

Hint: Count from end, remove that node [OK]

Common Mistakes:

• Removing wrong node counting from start
• Removing node 3 instead of 4
• Confusing 2nd from end with 2nd from start

4. Given this TypeScript code snippet to remove the nth node from end, what is the bug?

function removeNthFromEnd(head: ListNode | null, n: number): ListNode | null {
  let fast = head;
  let slow = head;
  for (let i = 0; i < n; i++) {
    fast = fast.next;
  }
  while (fast.next !== null) {
    fast = fast.next;
    slow = slow.next;
  }
  slow.next = slow.next.next;
  return head;
}

medium

A. It does not handle the case when n equals the length of the list

B. The loop runs one time too many causing runtime error

C. The slow pointer moves too fast and skips nodes

D. It returns the wrong head node

Solution

1. Step 1: Analyze fast pointer movement

   Fast pointer moves n steps ahead. If n equals list length, fast becomes null.

2. Step 2: Check while loop condition

   Code uses fast.next, but if fast is null, accessing fast.next causes error or skips removing head.

3. Step 3: Identify missing edge case handling

   Code does not handle removing the first node when n equals list length.

4. Final Answer:

   It does not handle the case when n equals the length of the list -> Option A

5. Quick Check:

   Edge case n = list length missing [OK]

Hint: Check if fast pointer is null after initial moves [OK]

Common Mistakes:

• Ignoring edge case when removing head node
• Assuming fast.next always exists
• Not returning updated head when head is removed

5. You have a linked list 1 -> 2 -> 3 -> 4 -> 5 -> null. You remove the 5th node from the end twice in a row. What is the final linked list?

hard

A. 2 -> 3 -> 4 -> 5 -> null

B. 3 -> 4 -> 5 -> null

C. 3 -> 4 -> null

D. 4 -> 5 -> null

Solution

1. Step 1: First removal of 5th node from end

   List length 5, 5th from end is head (1). After removal: 2 -> 3 -> 4 -> 5 -> null.

2. Step 2: Second removal of 5th node from end

   New length 4, n=5 > 4 so no 5th node exists; list unchanged: 2 -> 3 -> 4 -> 5 -> null.

3. Final Answer:

   2 -> 3 -> 4 -> 5 -> null -> Option A

4. Quick Check:

   Removing 5th from end twice removes head once [OK]

Hint: Removing nth from end twice may remove head once only [OK]

Common Mistakes:

• Confusing nth from end with nth from start
• Thinking the problem is about reversing the list
• Assuming insertion instead of deletion