Attentive·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

Coding round at Attentive for a software engineer role. One algorithmic problem, started with a constrained version then had to generalize it, which is a format I wasn't fully expecting.

Questions Asked (1)

Given a string and an integer k, repeatedly remove contiguous groups of exactly k identical characters, concatenating the remaining parts after each removal until no more removals are possible. Return the final string. Start by solving the case where k is fixed at 3, then generalize your solution to any k.

Algorithms & Data Structures

Author's notes

The warm-up framing threw me a bit.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and walking through a small example for k=3 to ensure understanding. Then propose an efficient stack-based solution that processes the string character by character, maintaining counts of consecutive identical characters, and popping when the count reaches k. Finally, discuss generalization to any k, complexity analysis, and edge cases.

Pro tip: Mention that a naive approach of repeatedly scanning and removing groups would be O(n^2) or worse, and that the stack-based method achieves O(n) time and space, which is optimal. Also, highlight that the stack can store pairs of (character, count) to avoid storing the entire string.

1. Clarify and Example

Restate the problem in your own words and walk through a small example for k=3, such as 'aabbbcc' -> 'aacc' -> 'aa' (no removal) or 'abbbaa' -> 'aaa' -> '' (if k=3). This ensures you and the interviewer agree on the rules.

2. Naive Approach and Inefficiency

Describe a straightforward approach: repeatedly scan the string, find groups of exactly k identical characters, remove them, and concatenate. Explain that this could be O(n^2) in the worst case due to multiple passes and string concatenations.

3. Optimal Stack-Based Solution

Propose using a stack where each element is a pair (character, count). Iterate through the string: if the current character matches the top, increment its count; otherwise, push (char, 1). If the count reaches k, pop the element. Finally, reconstruct the string from the stack.

4. Generalize to Any k

Explain that the same algorithm works for any k by simply using k as the threshold for popping. Discuss that k is a parameter and the logic remains unchanged.

5. Complexity and Edge Cases

Analyze time and space complexity: O(n) time and O(n) space in the worst case. Mention edge cases: empty string, k <= 0 (invalid), k=1 (removes all characters), and strings with no removals.

Key Points to Mention

Stack-based approach with (character, count) pairs for O(n) time and space.
Handling of concatenation implicitly by the stack, avoiding explicit string concatenation.
Generalization to any k by parameterizing the removal threshold.
Edge cases: empty string, k=1, k > string length, and no removals possible.
Complexity analysis: O(n) time and O(n) space, which is optimal.
Potential follow-up: if k is large, the stack size is bounded by n, but still O(n).

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