How do you remove the nth node from the end of a linked list?

Q: How do you remove the nth node from the end of a linked list?

**Target Audience:** Beginners preparing for coding interviews — you should know what a linked list node is; everything else is explained **Difficulty:** Medium **Language:** Python 3 > **How to read this doc:** Every idea is explained in plain language first, then tagged with `📌 Interview term:` so you also learn the exact word an interviewer expects. All code is Python, and every output block in the **Verified** section was copied from a real run of the examples on this page. ## 1. Why This Even Matters — A Story First You are in a queue of people at a ticket office and you are told: "the second person from the back has to leave". Nobody in the queue knows how long it is, and you can only walk from the front. The obvious plan is to walk to the back counting heads, then walk again from the front to the right spot. That works, but it takes two trips. A neater plan uses two friends and a rope. Tie a rope of a fixed length between them, so the front friend is always a fixed number o

The problem

Question presented to candidate: "Given the head of a singly linked list, remove the nth node from the end and return the head. For 1, 2, 3, 4, 5 and n = 2 the answer is 1, 2, 3, 5. Can you do it in one pass? Please write it in Python."

What a strong answer should cover:

The two-pass baseline: count the length, then unlink the node at index length - n.
The one-pass version: a dummy node, fast moved n + 1 links ahead, then both pointers move until fast is None; slow is the predecessor of the target.
Why the dummy node matters: removing the head (n = length) needs no special case.
O(L) time and O(1) space for both, and an honest note that "one pass" does not mean fewer pointer moves.
Python specifics: ListNode(0, head) as the dummy, for _ in range(n + 1), and the unlinked node is reclaimed by the garbage collector once nothing refers to it.

Clarifying questions expected:

"Is n 1-based and always between 1 and the length?"
"Should I return the new head?"

Code / implementation expected: Yes — remove_nth_from_end(head, n) in Python, one pass.

linked-list
two-pointers

Try it yourself

Sketch your approach before you read the solution. That is what the interview tests.

Stuck? Get a hint.

A short nudge toward the approach. It will not give the answer away.

Solution

Target Audience: Beginners preparing for coding interviews — you should know what a linked list node is; everything else is explained Difficulty: Medium Language: Python 3

How to read this doc: Every idea is explained in plain language first, then tagged with 📌 Interview term: so you also learn the exact word an interviewer expects. All code is Python, and every output block in the Verified section was copied from a real run of the examples on this page.

1. Why This Even Matters — A Story First

You are in a queue of people at a ticket office and you are told: "the second person from the back has to leave". Nobody in the queue knows how long it is, and you can only walk from the front. The obvious plan is to walk to the back counting heads, then walk again from the front to the right spot. That works, but it takes two trips.

A neater plan uses two friends a

The solution is hidden so you can try first

9 min read · 10 sections

Python code

Edit the code and run it here. Each printed line with an expected value becomes a test.

PythonOne-pass removal with a dummy node, including the edge cases

Practice

PythonOne pass versus two passes: same results, and the pointer moves counted

Practice

Discussion

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