Prime factorization
Given an integer n (1 <= n <= 100000), return its prime factorization as a list of [prime, exponent] pairs sorted by ascending prime. For example 360 = 2^3 * 3^2 * 5^1 returns [[2,3],[3,2],[5,1]]. Build a smallest-prime-factor sieve so the factorization itself runs in time proportional to the number of prime factors. For n = 1 return an empty list.
Implement
prime_factorization(n: int) → list[list[int]]Examples
in
[12]out[[2,2],[3,1]]What 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 25 min
solution.py
InputExpectedGot
[12][[2,2],[3,1]]not run yetsample