Remove Nth Node from End of List in DSA Typescript - Cheat Sheet & Quick Revision

Recall & Review

beginner

What is the main idea behind removing the nth node from the end of a linked list?

Use two pointers spaced n nodes apart. Move both until the first reaches the end. The second pointer will then be just before the node to remove.

beginner

Why do we use a dummy node at the start of the linked list in this problem?

A dummy node helps handle edge cases like removing the first node easily by providing a fixed node before the head.

intermediate

In the two-pointer approach, what happens if the first pointer moves n steps ahead and reaches null immediately?

It means n equals the length of the list, so the node to remove is the head node.

beginner

What is the time complexity of removing the nth node from the end of a linked list using the two-pointer method?

O(L), where L is the length of the linked list, because we traverse the list at most once.

beginner

How do you update the links to remove the nth node from the end once the second pointer is just before it?

Set second.next = second.next.next to skip the nth node and remove it from the list.

What does the first pointer do in the two-pointer approach to remove the nth node from the end?

A Moves to the end immediately

B Moves n steps ahead to create a gap

C Stays at the head

D Moves backwards

The first pointer moves n steps ahead to create a gap between the two pointers.

Why is a dummy node used before the head in this problem?

A To point to the last node

B To store the length of the list

C To simplify removing the head node

D To reverse the list

A dummy node simplifies edge cases like removing the head node by providing a stable node before the head.

If n equals the length of the list, which node is removed?

A The head node

B The last node

C The middle node

D No node is removed

If n equals the list length, the node to remove is the head node.

What is the time complexity of the two-pointer solution?

A O(1)

B O(n^2)

C O(log n)

D O(n)

The solution traverses the list at most once, so time complexity is O(n).

How do you remove the nth node once the second pointer is just before it?

A Set second.next to second.next.next

B Swap values with next node

C Move second pointer forward

D Set second.next to null

Setting second.next to second.next.next skips the nth node, effectively removing it.

Explain step-by-step how to remove the nth node from the end of a singly linked list using two pointers.

Think about creating a gap and moving pointers together.

Describe why a dummy node is helpful in removing the nth node from the end of a linked list.

Consider what happens if the head node needs to be removed.

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