Code Room
CodingMedium
Question
Return the smallest prime factor of a positive integer n (with n up to about 10^12). If n is prime, the smallest prime factor is n itself. Return -1 for n < 2. Handle the factor 2 first, then trial-divide by odd numbers up to sqrt(n); the first odd divisor found is the smallest prime factor.
Implement
smallest_prime_factor(n: int) → intExamples
in
[21]out3What a strong answer looks like
State your approach and its time/space complexity out loud before you optimize. Handle the edge cases (empty input, duplicates, overflow), and say why you chose this over the brute force. Green tests are the floor, not the grade.
Learn the concepts
Vibe coding: describe the solution in plain language (or narrate it) and the coach grades your approach. Generating runnable code from your description is coming next.
Run or narrate your approach, then ask the coach.