Grammarly·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Apr 2026Remote

Summary

Grammarly coding interview with a string manipulation problem that looks deceptively simple but has a neat stack-based trick to it. Nothing too wild, just one focused algorithmic question.

Questions Asked (1)

Given a string and an integer k, repeatedly remove groups of k adjacent identical characters until no more removals are possible. Return the resulting string. Solve it efficiently.

Algorithms & Data Structures

Author's notes

The naive approach of scanning and removing in a loop will get you killed on time complexity.

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AI HintsAI Generated

Suggested Approach

Clarify the problem and edge cases, then propose a stack-based solution that tracks character counts to efficiently remove groups of k identical characters. Walk through an example to demonstrate correctness and analyze time and space complexity.

Pro tip: Mention that the stack approach simulates the removal process in one pass, and highlight how it avoids repeated string scans, achieving O(n) time. Also, discuss potential follow-ups like handling Unicode or streaming input.

1. Clarify requirements and edge cases

Confirm that k is at least 2, the string can be empty, and removals can cascade. Ask about input size and character set to guide optimization.

2. Propose an efficient approach

Suggest using a stack of (character, count) pairs. Iterate through the string, pushing or incrementing counts, and popping when count reaches k.

3. Walk through an example

Trace the algorithm on a small example like 'deeedbbcccbdaa' with k=3 to show how removals cascade and the final string is built.

4. Analyze complexity

State that the time complexity is O(n) since each character is processed once, and space complexity is O(n) for the stack in the worst case.

5. Discuss optimizations and edge cases

Mention that the stack can be implemented with arrays for speed, and consider edge cases like k larger than string length or all characters identical.

Key Points to Mention

Stack-based approach with character counts
Single pass O(n) time complexity
Cascading removals handled by stack
Space complexity O(n) worst case
Edge cases: empty string, k > length, no removals
Comparison with naive repeated scanning approach

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