← Microsoft Interview Insights

Microsoft·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Apr 2026

Summary

Microsoft SWE coding round, one question on linked lists. Pretty standard but the O(1) space constraint is where things get interesting if you haven't thought about it before.

Questions Asked (1)

Q1

Given the head of a singly linked list, determine whether it forms a palindrome. Can you do it in O(n) time with O(1) extra space?

Algorithms & Data Structures
Author's notes

The naive approach with a stack or array is obvious and I think they expected me to start there, but the real question is the in-place version.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

First, clarify the constraints and confirm that O(1) space means we cannot use a stack or array. Then, describe the optimal approach: find the middle of the list using slow and fast pointers, reverse the second half in-place, compare the two halves, and finally restore the list to its original order. This achieves O(n) time and O(1) space.

Pro tip: Mention that you would restore the list to its original state after checking, as good practice to avoid side effects, and discuss edge cases like empty list, single node, and even/odd lengths.

1. Clarify requirements and constraints

Confirm that the list is singly linked, and that O(1) extra space means we cannot use additional data structures like arrays or stacks. Also, clarify if modifying the list is allowed (usually yes, but should be restored).

2. Find the middle of the linked list

Use the slow and fast pointer technique: slow moves one step, fast moves two steps. When fast reaches the end, slow is at the middle. For even length, slow will be at the start of the second half.

3. Reverse the second half

Reverse the linked list starting from the slow pointer (or slow.next for odd length) to the end. This can be done iteratively with three pointers: prev, current, and next.

4. Compare the two halves

Traverse from the head and from the reversed second half simultaneously, comparing node values. If all match, it's a palindrome; otherwise, it's not.

5. Restore the list and return result

Reverse the second half again to restore the original list structure, then return the boolean result. This step is optional but recommended for good practice.

Key Points to Mention

  • Time complexity: O(n) because we traverse the list a constant number of times (find middle, reverse, compare, restore).
  • Space complexity: O(1) because we only use a few pointers, no additional data structures.
  • Handling edge cases: empty list, single node, two nodes, and lists with odd/even number of nodes.
  • The slow and fast pointer technique for finding the middle.
  • In-place reversal of a linked list using iterative pointer manipulation.
  • Restoring the original list to avoid side effects, demonstrating attention to detail.

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