Grammarly·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Apr 2026

Summary

Grammarly coding screen, one question, stack-based string manipulation. Pretty clean problem once you see the pattern but I fumbled around for a bit before the approach clicked.

Questions Asked (1)

Given a string of lowercase letters, repeatedly remove pairs of adjacent identical characters until no more removals are possible. Return the final string. Solve it in O(n) using a stack.

Algorithms & Data Structures

Author's notes

Took me a second to realize brute-forcing it with repeated passes would be too slow.

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AI HintsAI Generated

Suggested Approach

Use a stack to process the string character by character. For each character, if the stack is not empty and the top equals the current character, pop the stack; otherwise, push the character. Finally, the stack contains the result in order.

Pro tip: After presenting the stack solution, mention that the same logic can be implemented using a StringBuilder as a stack to avoid extra space for the stack object, which is a common optimization in production code.

1. Understand the problem

Clarify that removals are repeated until no adjacent identical pairs remain, and that the process may create new pairs. Confirm that the solution must run in O(n) time.

2. Choose the right data structure

Select a stack because it efficiently handles the last-in-first-out nature of adjacent pair removal. Explain that the stack will hold characters that have not been removed yet.

3. Iterate and process each character

Loop through the string. For each character, compare it with the top of the stack. If they match, pop the stack; otherwise, push the character.

4. Construct the final string

After processing all characters, the stack contains the remaining characters in order. Convert the stack to a string (e.g., by joining its elements) and return it.

5. Analyze complexity and edge cases

State that time complexity is O(n) because each character is pushed and popped at most once, and space complexity is O(n) in the worst case. Discuss edge cases like empty string, all characters removable, and no removals.

Key Points to Mention

Stack-based approach for O(n) time complexity
Each character is processed once, leading to linear time
Space complexity is O(n) due to the stack
Handling of edge cases: empty string, string with all pairs, no pairs
Comparison with naive repeated removal approach (O(n^2))
Potential optimization: using a StringBuilder as a stack to reduce overhead

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