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Oracle·Software Engineer·Technical Phone Screen·Intermediate

IntermediatePrefer not to say
Jun 2026

Summary

Oracle SWE coding round, pretty standard stuff. One algorithm question and that was basically it.

Questions Asked (1)

Q1

Given a string containing only bracket characters ('(', ')', '{', '}', '[', ']'), write a function to determine whether the string is valid. A string is valid if every opening bracket is closed by the same type of bracket and in the correct order.

Algorithms & Data Structures
Author's notes

Classic stack problem.

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AI HintsAI Generated

Suggested Approach

Use a stack to track opening brackets. Iterate through the string: push opening brackets onto the stack; for closing brackets, check if the stack is non-empty and the top matches the corresponding opening bracket. At the end, the stack must be empty for the string to be valid.

Pro tip: Clarify edge cases upfront (empty string, odd length, non-bracket characters) and mention that early termination on odd length or mismatched closing bracket improves efficiency. Also, discuss time and space complexity: O(n) time and O(n) space in the worst case.

1. Understand the problem and edge cases

Restate the problem to ensure clarity. Identify edge cases: empty string (valid), odd length (invalid), strings with only opening or only closing brackets, and strings with mixed bracket types.

2. Choose the right data structure

Select a stack (LIFO) to track opening brackets. Explain why a stack is ideal: the most recent opening bracket must be closed first, matching the LIFO principle.

3. Design the algorithm

Iterate through each character. If it's an opening bracket, push it onto the stack. If it's a closing bracket, check if the stack is empty or if the top doesn't match the corresponding opening bracket; if so, return false. Otherwise, pop the stack.

4. Finalize and analyze

After iteration, return true only if the stack is empty. Analyze time complexity O(n) and space complexity O(n) in the worst case. Mention potential optimizations like early exit on odd length.

Key Points to Mention

  • Use a stack to track opening brackets.
  • Map closing brackets to their corresponding opening brackets for easy comparison.
  • Check for empty stack when encountering a closing bracket.
  • Ensure the stack is empty after processing all characters.
  • Time complexity: O(n), space complexity: O(n).
  • Handle edge cases: empty string, odd length, and non-bracket characters (if allowed).

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