Remove Nth Node from End of List in DSA Typescript - Time & Space Complexity

Time Complexity: Remove Nth Node from End of List

O(n)

Understanding Time Complexity

We want to understand how the time needed changes when we remove the nth node from the end of a linked list.

How does the work grow as the list gets longer?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

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

This code removes the nth node from the end by using two pointers moving through the list.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Two loops moving pointers through the list.
• How many times: The first loop runs n times, the second loop runs (length - n) times.

How Execution Grows With Input

As the list gets longer, the total steps grow roughly with the list size.

Input Size (n)	Approx. Operations
10	About 10 steps
100	About 100 steps
1000	About 1000 steps

Pattern observation: The work grows linearly as the list size increases.

Final Time Complexity

Time Complexity: O(n)

This means the time to remove the node grows in direct proportion to the list length.

Common Mistake

[X] Wrong: "Since we only move pointers a few times, the time is constant O(1)."

[OK] Correct: The pointers must move through the list nodes, which depends on the list length, so time grows with n.

Interview Connect

Understanding this helps you handle linked list problems efficiently, a common skill interviewers look for.

Self-Check

"What if we used recursion instead of two pointers? How would the time complexity change?"

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