← Whatnot Interview Insights

Whatnot·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jun 2026

Summary

Did a coding round for a Software Engineer role at Whatnot. Just the one question, stack-based string problem, nothing too wild but worth knowing cold.

Questions Asked (1)

Q1

Given a string, repeatedly remove adjacent duplicate characters until no more adjacent duplicates exist. Return the final string.

Algorithms & Data Structures
Author's notes

Classic stack problem once you see it, but I fumbled around for a minute trying to think of a simulation approach before the stack click happened.

Create a free account to read the full note

AI HintsAI Generated

Suggested Approach

Use a stack to process the string character by character: if the current character matches the top of the stack, pop it; otherwise, push it. This naturally handles cascading removals in a single pass, yielding O(n) time and O(n) space.

Pro tip: After presenting the stack solution, mention that a two-pointer in-place approach can achieve O(1) extra space if the input is a mutable array, showing you optimize for memory when appropriate.

1. Clarify and Confirm

Ask about input constraints (e.g., character set, string length, mutability) and confirm that removal is case-sensitive and applies repeatedly until no adjacent duplicates remain.

2. Explain the Stack Approach

Describe iterating through the string and using a stack: push if different from top, pop if same. This simulates the repeated removal process in one pass.

3. Walk Through an Example

Trace the algorithm on a sample like 'abbaca' to demonstrate how cascading removals happen (e.g., 'bb' removed, then 'aa' removed, leaving 'ca').

4. Analyze Complexity

State that time complexity is O(n) because each character is pushed and popped at most once, and space complexity is O(n) for the stack in the worst case.

5. Discuss Optimizations and Edge Cases

Mention the two-pointer in-place variant for O(1) extra space, and handle edge cases like empty string, all duplicates, and no duplicates.

Key Points to Mention

  • Stack-based single-pass solution
  • Time complexity O(n) and space complexity O(n)
  • Cascading removals handled automatically by stack
  • Two-pointer in-place optimization for O(1) extra space
  • Edge cases: empty string, all same characters, no duplicates
  • Comparison with naive repeated scanning approach (O(n^2))

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