Bytedance·Machine Learning Engineer·Technical Phone Screen·Senior

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Jun 2026

Summary

Bytedance ML engineer interview that was basically a straight-up linked list coding session. The question was a twist on a classic hard problem and I ran out of time before I could even write tests, which felt bad.

Questions Asked (1)

Reverse nodes in a linked list in groups of k, but unlike the standard version, any remaining nodes at the end that don't fill a complete group should also be reversed. You also need to define the ListNode structure yourself.

Algorithms & Data Structures

Author's notes

The base problem I knew, but the variation tripped me up a bit.

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AI HintsAI Generated

Suggested Approach

First, define the ListNode class with val and next attributes. Then, use an iterative approach with a dummy node to reverse each group of k nodes, including the final partial group. Maintain pointers to the previous group's end and the current group's start, reversing the group by adjusting next pointers.

Pro tip: Clarify with the interviewer whether k can be 1 or 0, and handle edge cases like empty list or k <= 1 by returning the head unchanged. Also, consider using a recursive approach for cleaner code, but be prepared to discuss its space complexity.

1. Define ListNode and handle edge cases

Define a ListNode class with val and next. Check if head is None or k <= 1; if so, return head immediately.

2. Set up dummy node and pointers

Create a dummy node pointing to head. Use pointers: group_prev (node before current group), and a helper to reverse k nodes.

3. Reverse each group iteratively

For each group, find the kth node (or end of list). Reverse the group by adjusting next pointers, then connect the reversed group back to group_prev and the next group.

4. Handle final partial group

After the loop, if there are remaining nodes (less than k), reverse them as well and connect appropriately.

5. Return the new head

Return dummy.next as the new head of the modified list.

Key Points to Mention

Definition of ListNode with val and next.
Use of dummy node to simplify edge cases.
Iterative reversal of k nodes using three pointers (prev, curr, next).
Handling of final partial group by reversing remaining nodes.
Time complexity O(n) and space complexity O(1).
Edge cases: empty list, k=1, k greater than list length.

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