Voleon·Software Engineer·Technical Phone Screen·Intermediate

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Jul 2026Remote

Summary

Voleon threw a recursive string parsing problem at me that looked deceptively clean on the surface. The constraint to hit O(n) was the real test.

Questions Asked (1)

A 'good string' is defined recursively: '0' is a good string, and if A and B are both good strings, then '1' + A + B is also a good string. Given a binary string, determine whether it is a good string. Your solution must run in O(n) time.

Algorithms & Data Structures

Author's notes

My first instinct was to write a recursive parser and call it a day, which technically works but I fumbled explaining the O(n) guarantee because I kept second-guessing whether the recursion would revisit characters.

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AI HintsAI Generated

Suggested Approach

First, clarify the recursive definition and derive an iterative characterization: a good string is either '0' or '1' followed by two good strings. Then, design an O(n) algorithm using a stack or a counter to validate the structure, ensuring that every '1' has exactly two good substrings following it.

Pro tip: Emphasize that the grammar is unambiguous and can be parsed deterministically in one pass, which is key to achieving O(n) time. Also, mention that the number of '1's must be one less than the number of '0's for a valid good string, as a quick sanity check.

1. Understand the recursive definition

Restate the definition: '0' is good; if A and B are good, then '1' + A + B is good. This implies a binary tree structure where each '1' node has exactly two children that are good strings.

2. Derive an iterative parsing strategy

Use a stack to simulate the recursive parsing. Scan the string from left to right: push '0' as a completed good string; when encountering '1', ensure there are at least two good strings on the stack to combine into a new good string.

3. Implement the O(n) algorithm

Initialize a counter (or stack) to track the number of completed good strings. For each character: if '0', increment counter; if '1', check if counter >= 2, then decrement counter by 1 (since two good strings combine into one). At the end, the string is good if counter == 1 and no invalid operations occurred.

4. Validate with edge cases

Test with simple cases: '0' (good), '100' (good: 1 + '0' + '0'), '11000' (good: 1 + '10' + '0'? Actually '10' is not good; correct example: '11000' is not good). Also test invalid strings like '1', '01', '111000' to ensure the algorithm correctly rejects them.

5. Analyze time and space complexity

The algorithm runs in O(n) time with a single pass. Space can be O(1) if using a counter, or O(n) if using a stack, but a counter suffices because we only need the count of completed good strings.

Key Points to Mention

The recursive definition implies a full binary tree structure where each internal node is '1' and leaves are '0'.
A necessary condition: number of '1's = number of '0's - 1, but not sufficient.
The parsing can be done in one pass using a stack or a counter, achieving O(n) time.
The grammar is unambiguous, so deterministic parsing is possible.
Edge cases: empty string is not good; single '0' is good; strings with insufficient '0's are invalid.
Space optimization: use a counter instead of a stack to achieve O(1) space.

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