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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
May 2026Remote

Summary

Bytedance SWE interview that threw a classic LC problem at me and then immediately asked me to extend it in a way I hadn't prepped for. The follow-up was the real test.

Questions Asked (2)

Q1

Given a string containing '(', ')', and '*' where '*' can represent '(', ')', or an empty string, determine whether the string can be made into a valid parenthesis sequence.

Algorithms & Data Structures
Author's notes

I knew this one.

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AI HintsAI Generated

Suggested Approach

Use a greedy approach with two passes: first left-to-right treating '*' as '(' to ensure no prefix has more ')' than available '(' and '*', then right-to-left treating '*' as ')' to ensure no suffix has more '(' than available ')' and '*'. Alternatively, maintain a range of possible open counts (minOpen, maxOpen) in one pass, ensuring minOpen never negative and maxOpen non-negative at end.

Pro tip: Clarify with the interviewer whether '*' can be chosen independently for each occurrence, and mention that the range method is more space-efficient and can be done in one pass.

1. Understand the problem

Restate the problem: given a string with '(', ')', and '*', where '*' can be '(', ')', or empty, determine if it can be a valid parentheses sequence. Confirm that '*' choices are independent.

2. Choose an approach

Decide between two-pass greedy or single-pass range method. Explain the chosen approach and why it works.

3. Implement the algorithm

For two-pass: traverse left-to-right, count balance treating '*' as '('; if balance negative, return false. Then traverse right-to-left, treating '*' as ')'; if balance negative, return false. For range method: maintain minOpen and maxOpen, update based on character, ensure minOpen >= 0, and at end minOpen == 0.

4. Analyze complexity

State time complexity O(n) and space complexity O(1) for both approaches.

5. Test with examples

Walk through examples like '()', '(*)', '(*))', '((*' to verify correctness and edge cases.

Key Points to Mention

  • Greedy strategy: '*' can be used flexibly to balance parentheses.
  • Two-pass approach: left-to-right ensures no excess ')', right-to-left ensures no excess '('.
  • Range method: maintain minOpen and maxOpen to represent possible open counts.
  • Time complexity O(n), space complexity O(1).
  • Edge cases: empty string, string with only '*', strings with odd length (immediately false).
  • Clarify assumptions: '*' can be chosen independently for each occurrence.

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

Q2

Follow-up: using DFS or backtracking, enumerate ALL distinct valid strings you can produce by replacing each '*' with '(', ')', or an empty string. Write your own test cases and verify them.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

This is where I started sweating.

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AI HintsAI Generated

Suggested Approach

Clarify the problem: given a string with '*' wildcards, replace each '*' with '(', ')', or empty string, and enumerate all distinct valid strings. Use DFS/backtracking to explore all possibilities, pruning branches that cannot lead to valid strings (e.g., negative balance or impossible to close). After generating, deduplicate results and test with custom cases including edge cases like empty string, no '*', and multiple '*'.

Pro tip: Demonstrate strong testing discipline by writing test cases that cover edge cases (empty string, all '*', unbalanced parentheses) and verifying outputs, including checking for duplicates and validity. This shows you think about correctness and robustness, not just the algorithm.

1. Clarify requirements and constraints

Confirm what 'valid' means (balanced parentheses, possibly empty string), whether the input can contain other characters, and if the output should be sorted or deduplicated. Ask about input size to discuss complexity.

2. Design DFS/backtracking with pruning

At each '*', branch into three choices: '(', ')', or skip. Maintain a balance counter (open minus close) and prune if balance < 0 or if remaining characters cannot possibly balance (e.g., balance > remaining length).

3. Implement and deduplicate

Recursively build the string. When reaching the end, if balance == 0, add to a set to avoid duplicates. Alternatively, sort and skip duplicates during recursion if input has repeated patterns.

4. Write test cases and verify

Create tests: empty string, no '*', single '*', multiple '*', and cases with existing parentheses. Manually compute expected outputs for small cases and compare. Also test performance for larger inputs.

5. Analyze complexity and trade-offs

Discuss time complexity: O(3^k) where k is number of '*', but pruning reduces it. Space: O(k) recursion depth plus output size. Mention trade-offs between deduplication methods (set vs. sorting).

Key Points to Mention

  • Backtracking with pruning based on balance and remaining characters to avoid invalid branches.
  • Deduplication strategy: using a set to store results or sorting and skipping duplicates during recursion.
  • Validity condition: balance must be zero at the end and never negative during construction.
  • Edge cases: empty string, no '*', multiple '*', and strings with existing parentheses.
  • Complexity analysis: worst-case O(3^k) time, but pruning improves average case; space O(k) for recursion.
  • Testing methodology: write small cases, manually verify, and include stress tests for performance.

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