Prime exponent in factorial
Given a number n and a prime p, return the exponent of p in the prime factorization of n! (n factorial) using Legendre's formula: sum over k>=1 of floor(n / p^k). Do NOT compute n! itself (it would overflow instantly). For p=5 this also equals the number of trailing zeros of n! when paired with the (larger) power of 2. Constraints: 0 <= n <= 10^9, p is a prime <= 10^9.
Implement
prime_power_in_factorial(n: int, p: int) → intExamples
in
[100,5]out24What 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.
0:00 of about 22 min
solution.py
InputExpectedGot
[100,5]24not run yetsample