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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Bytedance SWE interview, second half was a linked list problem. Pretty standard but the k-group constraint adds enough complexity to trip you up if you're not careful.

Questions Asked (1)

Q1

Given a linked list and an integer k, reverse the nodes of the list in groups of k and return the modified list.

Algorithms & Data Structures
Author's notes

The base reversal is fine but the grouping logic is where things get messy.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints (e.g., whether to reverse leftover nodes if fewer than k remain) and then walk through an iterative solution using a dummy node to handle edge cases. Explain the process of reversing each group of k nodes by adjusting pointers, and analyze time and space complexity.

Pro tip: Demonstrate thoroughness by discussing edge cases upfront (k=1, k greater than list length, empty list) and mentioning that you can solve it iteratively in O(n) time and O(1) space, which is optimal.

1. Clarify requirements and edge cases

Ask if leftover nodes (fewer than k) should remain as is or be reversed. Confirm assumptions about k (e.g., k >= 1) and handle edge cases like empty list, k=1, and k > list length.

2. Choose an approach

Decide between iterative and recursive solutions. For interviews, an iterative approach with O(1) space is often preferred, but mention recursion as an alternative with O(n/k) stack space.

3. Outline the algorithm

Use a dummy node pointing to the head. For each group of k nodes, reverse the pointers within the group and connect the previous group's tail to the new head. Keep track of the next group's start.

4. Walk through an example

Trace the algorithm on a small example (e.g., 1->2->3->4->5, k=2) to demonstrate correctness and pointer manipulation.

5. Analyze complexity and test

State time complexity O(n) and space O(1). Discuss potential pitfalls like losing references and how to avoid them. Mention testing with edge cases.

Key Points to Mention

  • Use of a dummy node to simplify edge cases and return the new head.
  • Pointer manipulation: reversing links within each group and connecting groups.
  • Handling of leftover nodes when the list length is not a multiple of k.
  • Time complexity O(n) and space complexity O(1) for iterative solution.
  • Edge cases: empty list, k=1, k > list length, and k=0 (if allowed).
  • Comparison with recursive approach and its space complexity.

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