Remove Nth Node From End of List

mediumTests passed 2026-11-11v1.0.0

TC: O(N)|SC: O(1)

Linked Lists Original Statement (leetcode).md

1. Core Invariant / The Trick

Attach a `dummy` before `head`. Advance `right` `n + 1` steps ahead of `left = dummy`, or advance `right` `n` steps then step both until `right.next == null` so `left` lands cleanly on the predecessor.

2. Complexity Bounds & Proof

Single one-pass scan preserving exact `n`-node separation between `left` and `right`.

3. Minimal Idiomatic Code

def removeNthFromEnd(head: Optional[ListNode], n: int) -> Optional[ListNode]:
    dummy = ListNode(0, head)
    left = right = dummy
    for _ in range(n):
        right = right.next
    while right.next:
        left = left.next
        right = right.next
    left.next = left.next.next
    return dummy.next

4. Failure Modes & Interview Follow-ups

Pitfalls:- Deleting the head node (`n == length`) crashes without a dummy node before `head`. - Off-by-one step count stops `left` on the target node instead of its predecessor.

Follow-ups:- Delete the Middle Node of a Linked List uses `dummy` + fast/slow pointers with the same predecessor pattern.

Personal Notes & Scratchpad (Local-First Yjs CRDT)

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Related Invariants

Reverse Linked List (O(N))

Browser Scratchpad

Console Output

Click Run or press Cmd/Ctrl+Enter to execute locally in a Web Worker.