#19 - Remove Nth Node From End of List

MediumLinked List

Given the head of a linked list, remove the n-th node from the end of the list and return its head.

Input	Output
head = [1,2,3,4,5], n = 2	[1,2,3,5]
head = [1], n = 1	[]
head = [1,2], n = 1	[1]

Solution Space

Try it

/**
 * Removes the n-th node from the end of the list in one pass.
 * Uses a dummy node to simplify edge cases (e.g. removing the head).
 *
 * @param head Head of the linked list
 * @param n    1-based position from the end to remove
 * @returns    Head of the modified list
 */
export function removeNthFromEnd(
  head: ListNode | null,
  n: number,
): ListNode | null {
  const dummy = new ListNode(0, head);
  let first: ListNode | null = dummy;
  let second: ListNode | null = dummy;

  // Advance first pointer n + 1 steps so the gap is n nodes
  for (let i = 0; i <= n; i++) {
    first = first!.next;
  }

  // Move both until first reaches the end
  while (first) {
    first = first.next;
    second = second!.next;
  }

  // second is now right before the target node — skip it
  second!.next = second!.next!.next;

  return dummy.next;
}

Approach: two-pointer gap technique – create a gap of `n` nodes between `first` and `second`. When `first` reaches the end, `second` sits right before the node to remove. A dummy node before `head` handles the edge case of removing the head itself.

Time: O(L) where L is the list length – single pass.

Space: O(1) – only pointer variables and one dummy node.

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