Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say

Jun 2026

Summary

Bytedance coding screen, one question on parentheses generation. Pretty standard stuff but the implementation details trip you up if you haven't drilled backtracking recently.

Questions Asked (1)

Given an integer n, generate all valid combinations of n pairs of parentheses.

Algorithms & Data Structures

Author's notes

I knew it was backtracking pretty fast but fumbled explaining the pruning conditions out loud.

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AI HintsAI Generated

Suggested Approach

Use backtracking to build the parentheses string incrementally, ensuring at each step that the number of closing parentheses never exceeds the number of opening ones and that we never exceed n pairs. Recursively explore adding an opening parenthesis if we have remaining opens, and a closing parenthesis if closes < opens. Collect all valid strings when the length reaches 2n.

Pro tip: Mention that this is essentially generating all valid sequences with a balance constraint, and that the number of valid combinations is the nth Catalan number, which shows you understand the combinatorial structure and can discuss complexity in terms of Catalan numbers.

1. Clarify the problem and constraints

Confirm that n is a non-negative integer and that we need all distinct valid combinations. Discuss edge cases like n=0 (should return an empty list or a list with an empty string, depending on definition).

2. Define the backtracking state

Track the current string, the number of open parentheses used, and the number of close parentheses used. The recursion depth will be at most 2n.

3. Establish base case and recursive choices

When the current string length equals 2n, add it to the result. Otherwise, if open < n, add '(' and recurse; if close < open, add ')' and recurse.

4. Implement and analyze complexity

Code the backtracking function iteratively or recursively. Analyze time complexity as O(4^n / sqrt(n)) or O(C_n * n) where C_n is the nth Catalan number, and space complexity as O(n) for recursion stack plus output storage.

5. Test with examples and discuss optimizations

Walk through n=3 to show the generation order. Mention that pruning invalid branches early makes it efficient, and that no additional optimizations are typically needed.

Key Points to Mention

Backtracking with pruning: only add ')' if it won't make the string invalid.
The number of valid combinations is the nth Catalan number.
Time complexity is exponential but bounded by Catalan number, often expressed as O(4^n / sqrt(n)).
Space complexity is O(n) for recursion stack, plus O(C_n * n) for storing results.
Edge cases: n=0, n=1, and negative n (should handle or clarify).
Alternative approaches: dynamic programming or BFS, but backtracking is most intuitive.

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