Microsoft·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Microsoft SWE interview with a linked list problem that looks manageable until you're actually in it. The in-place constraint is where things get interesting.

Questions Asked (1)

Given the head of a singly linked list and an integer k, reverse the nodes in groups of k in-place. If the final group has fewer than k nodes, leave them as-is. Walk through your algorithm and implement it.

Algorithms & Data StructuresTechnical Trade-offs

Author's notes

I knew the rough idea pretty fast, reverse chunks of k, stitch them back together, move on.

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AI HintsAI Generated

Suggested Approach

Clarify the problem constraints and edge cases, then explain a pointer-based iterative approach that reverses each full group of k nodes in-place while preserving the remaining nodes. Walk through the algorithm with a small example, analyze time and space complexity, and finally implement it cleanly with careful pointer manipulation.

Pro tip: Emphasize that you check if there are at least k nodes remaining before reversing a group, and use a dummy node to simplify handling the new head. This shows attention to edge cases and clean code.

1. Clarify and Confirm

Restate the problem, ask about constraints (e.g., k=1, k greater than list length, empty list), and confirm in-place requirement and return value.

2. High-Level Approach

Explain that you will traverse the list, and for each group of k nodes, reverse the links within the group, then connect the previous group's tail to the new head of the reversed group.

3. Detailed Algorithm Walkthrough

Describe using a dummy node, pointers for group_prev, and a loop that checks for k nodes ahead. Reverse the group by adjusting next pointers, then update group_prev to the original group's first node (now tail).

4. Complexity Analysis

State that time complexity is O(n) since each node is visited once, and space complexity is O(1) as reversal is done in-place.

5. Implementation and Testing

Write clean code with meaningful variable names, then test with edge cases: empty list, k=1, k equal to length, k greater than length, and multiple groups.

Key Points to Mention

Use a dummy node to handle the case where the head changes, simplifying edge cases.
Before reversing a group, check if there are at least k nodes remaining; if not, leave the rest as-is.
During reversal, keep track of the node after the group (group_next) to reconnect after reversal.
After reversing a group, connect the previous group's tail to the new head, and the new tail to the next group's head.
Time complexity O(n) and space complexity O(1) due to in-place pointer manipulation.
Handle edge cases: empty list, k=1, k > list length, and k equal to list length.

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