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Bytedance·Software Engineer·Technical Phone Screen·Intermediate

Intermediate
Jul 2026

Summary

Bytedance SWE interview with a numerical methods problem that sounds easy until you actually think about the precision requirement. Pretty standard technical phone screen vibe, just one meaty coding question.

Questions Asked (1)

Q1

Given a non-negative integer val and a non-negative integer precise, compute the square root of val without using any built-in square root function. The result should have an absolute error of at most 10^-precise. Walk through the time complexity of your approach.

Algorithms & Data StructuresTechnical Trade-offs
Author's notes

Binary search on a float range was the move, but I kept second-guessing my termination condition.

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AI HintsAI Generated

Suggested Approach

Use Newton's method (or binary search) to iteratively approximate the square root, stopping when the change is less than 10^-precise. Then analyze the time complexity based on the number of iterations and the cost of arithmetic operations.

Pro tip: Mention that Newton's method converges quadratically, so the number of iterations is logarithmic in the required precision, making it efficient for high precision. Also, clarify how you handle edge cases like val=0 and precise=0.

1. Clarify requirements and edge cases

Confirm that val and precise are non-negative integers, and that the result must be a floating-point number with absolute error ≤ 10^-precise. Discuss handling val=0 and precise=0.

2. Choose an algorithm

Select Newton's method for quadratic convergence or binary search for simplicity. Explain the trade-offs: Newton's method is faster but requires careful initialization; binary search is robust but slower.

3. Implement the algorithm

For Newton's method, start with an initial guess (e.g., val or 1) and iterate x = (x + val/x)/2 until the change is less than 10^-precise. For binary search, set low=0, high=max(1, val), and narrow the interval until the width is less than 10^-precise.

4. Analyze time complexity

For Newton's method, the number of iterations is O(log(precise)) due to quadratic convergence, and each iteration involves constant-time arithmetic operations (assuming fixed-precision arithmetic). For binary search, the number of iterations is O(precise) because the interval halves each time, leading to O(precise) iterations.

5. Discuss precision and potential pitfalls

Address floating-point precision issues, such as when val is very large or precise is high. Mention using epsilon = 10^-precise for termination and ensuring the error bound is met.

Key Points to Mention

  • Newton's method converges quadratically, so iterations are O(log(precise)).
  • Binary search takes O(precise) iterations because the interval halves each time.
  • Each iteration involves basic arithmetic operations, assumed O(1).
  • Handle edge cases: val=0, precise=0, and large val.
  • Use epsilon = 10^-precise for termination condition.
  • Floating-point precision may affect the actual error; consider using higher precision if needed.

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