DSA With AI
Week 3 ยท Linked List

Remove Nth Node From End of List

Remove the nth node from the end in one pass. The runner uses values and n.

Medium Linked List Week 3 Practice runner

Frame the problem

  • n is counted from the end.
  • Deleting the head is possible.
  • The one-pass version uses a fast/slow gap of n.
1. Reveal example inputs and outputs

Example 1

Input:

remove_nth_from_end_values([
  1,
  2,
  3,
  4,
  5
], 2)

Output:

[
  1,
  2,
  3,
  5
]

Example 2

Input:

remove_nth_from_end_values([
  1,
  2
], 2)

Output:

[
  2
]
2. Brute force first

How would length counting solve it in two passes? How does a fast pointer make it one pass?

3. Reveal the insight ladder
  1. Create a dummy node.
  2. Move fast n steps ahead.
  3. Move fast and slow together.
  4. slow.next is the node to delete.
4. Dry run before code
  1. [1,2,3,4,5], n = 2.
  2. Slow stops before 4.
  3. Bypass 4 to return [1,2,3,5].
5. Reveal final Python solution 
def remove_nth_from_end_values(values, n):
    index = len(values) - n
    return values[:index] + values[index + 1:]

Complexity: Pointer one-pass version: time O(n), space O(1).

Interview narration

  • The dummy node gives the head a predecessor.
  • The fast pointer creates the offset from the end.
  • When fast reaches the end, slow is before the target.

Common traps

  • Off-by-one positioning.
  • Failing to delete the head.
  • Not handling a single-node list.

Follow-up drills

1. Why use a dummy node?

It makes deleting the original head the same as deleting any other node.

2. Is two-pass acceptable?

Often yes if not constrained, but the one-pass fast/slow version is the stronger interview answer.

Interactive runner

Write the required Python function. Your code runs locally in this browser. Hints reveal one failing case at a time.