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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Bytedance infrastructure round, one algorithmic question on linked lists. Pretty standard but they wanted a specific optimal approach so it wasn't just about getting the right answer.

Questions Asked (1)

Q1

Given the head of a singly linked list and an integer n, remove the nth node from the end of the list and return the head. Can you do it in a single pass with constant space?

Algorithms & Data Structures
Author's notes

The two-pointer approach is the whole point here, they weren't satisfied with anything that required two traversals.

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AI HintsAI Generated

Suggested Approach

Use the two-pointer technique: advance a fast pointer n+1 steps ahead, then move both pointers until fast reaches the end. The slow pointer will be just before the node to remove, allowing you to skip it in one pass with constant space. Handle edge cases like removing the head by using a dummy node.

Pro tip: Mention that using a dummy node simplifies edge cases, especially when the head needs to be removed, and always clarify the definition of 'n' (1-indexed from the end) and constraints (e.g., n is valid).

1. Clarify the problem and edge cases

Confirm that n is 1-indexed from the end, the list may have only one node, and n is always valid. Discuss handling removal of the head.

2. Choose the two-pointer approach

Explain that you'll use two pointers (fast and slow) to find the node to remove in one pass without knowing the list length.

3. Initialize with a dummy node

Create a dummy node pointing to the head to simplify edge cases, and set both pointers to the dummy node.

4. Advance fast pointer and move both

Move fast n+1 steps ahead, then move both pointers until fast reaches null. Slow will point to the node before the one to remove.

5. Remove the node and return

Skip the target node by updating slow.next, then return dummy.next as the new head.

Key Points to Mention

  • Two-pointer technique (fast and slow pointers) for single-pass traversal.
  • Use of a dummy node to handle edge cases like removing the head.
  • Time complexity O(L) where L is list length, space complexity O(1).
  • The gap between fast and slow pointers is n+1 to position slow before the target.
  • Handling of n=1 (removing the last node) and n=length (removing the head).
  • Clarify assumptions: n is valid, list is singly linked, and return the head of the modified list.

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.