Molocoads·Machine Learning Engineer·Technical Phone Screen·Senior

Senior

Jun 2026

Summary

Interviewed for an ML Engineer role at Molocoads and got hit with a pure algorithms problem. Stack-based string manipulation, nothing ML about it, but I guess they want to know you can actually code.

Questions Asked (1)

Given a string and an integer k, repeatedly remove groups of k adjacent identical characters until no such group exists. Return the resulting string. What's an O(n) solution?

Algorithms & Data Structures

Author's notes

The naive approach is obvious and they probably expect you to dismiss it quickly.

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AI HintsAI Generated

Suggested Approach

Use a stack to process characters one by one, maintaining counts of consecutive identical characters. When the count reaches k, pop the group from the stack. This yields an O(n) solution because each character is pushed and popped at most once.

Pro tip: After explaining the stack approach, mention that this is essentially a run-length encoding with a stack, and that the same technique can be used for similar problems like removing adjacent duplicates. Also, clarify that the removal can cascade, so you must continue checking after a pop.

1. Clarify the problem

Confirm that groups of exactly k adjacent identical characters are removed, and that removal can cause new groups to form, requiring repeated passes until no such group exists.

2. Identify the need for O(n)

Recognize that a naive simulation with multiple passes could be O(n^2) or worse, so we need a linear-time approach.

3. Design the stack-based solution

Use a stack where each element stores a character and its current consecutive count. For each character in the input, if it matches the top, increment the count; otherwise push with count 1. If the count reaches k, pop the element.

4. Analyze complexity

Explain that each character is pushed and popped at most once, so the time complexity is O(n) and space complexity is O(n) in the worst case.

5. Test with examples

Walk through a small example, such as s = 'deeedbbcccbdaa', k = 3, to demonstrate how the stack handles cascading removals.

Key Points to Mention

Stack stores pairs of (character, count) to avoid rescanning.
When count equals k, pop the group, which may cause the new top to merge with subsequent characters.
The algorithm processes each character exactly once, ensuring O(n) time.
Space complexity is O(n) due to the stack, but can be O(n/k) in some cases? Actually worst-case O(n).
Edge cases: empty string, k=1 (remove all characters), k > string length.
The solution handles cascading removals naturally because after popping, the next character is compared with the new top.

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