Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate

May 2026

Summary

Bytedance SWE interview with a classic bracket validation problem. Pretty standard algorithmic round, nothing too surprising, but the stack constraint made it feel more intentional than a throwaway warmup.

Questions Asked (1)

Given a string made up only of parentheses, curly braces, and square brackets, write a function to determine if the string is valid. Valid means every opening bracket is closed by the matching type, and they close in the correct nested order. You must solve it using a stack in O(n) time.

Algorithms & Data Structures

Author's notes

I knew this one pretty quickly but still managed to fumble the edge cases early on.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the problem and edge cases, then explain the stack-based algorithm: iterate through the string, push opening brackets, and for closing brackets check if the top of the stack matches. Conclude with complexity analysis and test with examples.

Pro tip: Mention that using a stack is optimal because brackets follow LIFO order, and proactively discuss edge cases like empty string and early termination to show thoroughness.

1. Clarify and Confirm

Restate the problem to ensure understanding, and ask about edge cases like empty string or strings with only opening brackets.

2. Outline the Stack Approach

Explain that you'll use a stack to track opening brackets, and a hash map to match closing brackets to their corresponding openings.

3. Walk Through the Algorithm

Describe iterating through each character: if it's an opening bracket, push it; if it's a closing bracket, check if the stack is non-empty and the top matches, then pop; otherwise return false.

4. Handle Edge Cases and Final Check

After the loop, ensure the stack is empty; also mention early termination if a mismatch is found or if a closing bracket appears with an empty stack.

5. Analyze Complexity and Test

State that time complexity is O(n) and space complexity is O(n) in the worst case, then walk through a few test cases like '()[]{}', '([)]', and ''.

Key Points to Mention

Use a stack to enforce LIFO order for nested brackets.
Use a hash map for O(1) matching of closing brackets to opening brackets.
Check for empty stack when encountering a closing bracket.
After processing, the stack must be empty for the string to be valid.
Time complexity O(n) and space complexity O(n) due to stack.
Edge cases: empty string, single type, mismatched types, and early termination.

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