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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026

Summary

Bytedance SWE interview, coding round. One linked list problem that looks straightforward until you actually try to implement it cleanly under pressure.

Questions Asked (1)

Q1

Given the head of a singly linked list and an integer k, reverse the nodes in consecutive groups of k. If the final group has fewer than k nodes, leave it as-is. You must rewire pointers, not copy values, and ideally do it in O(1) extra space.

Algorithms & Data Structures
Author's notes

My first instinct was recursion and it worked, but they asked the follow-up about doing it without recursion or extra buffers and that's where I fumbled a bit.

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AI HintsAI Generated

Suggested Approach

Use an iterative approach with a dummy node to handle edge cases, reversing each group of k nodes by rewiring pointers. Track the node before each group and the node after, then reconnect the reversed group. If fewer than k nodes remain, leave them as-is.

Pro tip: Clarify whether k=1 or k greater than list length should be handled, and mention that you can first count the length to avoid unnecessary reversals. Also, emphasize that you're not using extra space beyond a few pointers, achieving O(1) space.

1. Understand the problem and edge cases

Confirm constraints: k >= 1, list may be empty, k may exceed list length. Clarify that only full groups of k are reversed.

2. Set up pointers and dummy node

Create a dummy node pointing to head to simplify edge cases. Use pointers: group_prev (node before current group), and a pointer to traverse k nodes.

3. Reverse each group of k nodes

For each group, check if there are at least k nodes. If so, reverse the k nodes by rewiring pointers, then connect group_prev to the new head of the reversed group and the new tail to the next group.

4. Handle the final incomplete group

After processing all full groups, if fewer than k nodes remain, leave them as-is and terminate.

5. Return the new head and analyze complexity

Return dummy.next as the new head. State time complexity O(n) and space complexity O(1).

Key Points to Mention

  • Use of dummy node to simplify edge cases (e.g., reversing from head).
  • Pointer manipulation: tracking group_prev, group_next, and reversing within group.
  • Checking for at least k nodes before reversing to handle incomplete final group.
  • Time complexity O(n) and space complexity O(1).
  • Avoid copying values; only rewire pointers.
  • Handling edge cases: empty list, k=1, k > list length.

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