he-025

# HumanEval/25

Implement a file **`solution.py`** that completes the function below. Keep the given name and signature; your file must define it at module level.

```python
from typing import List


def factorize(n: int) -> List[int]:
    """ Return list of prime factors of given integer in the order from smallest to largest.
    Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
    Input number should be equal to the product of all factors
    >>> factorize(8)
    [2, 2, 2]
    >>> factorize(25)
    [5, 5]
    >>> factorize(70)
    [2, 5, 7]
    """
```

<!-- imported from OpenAI HumanEval (HumanEval/25) -->

# Auto-generated from OpenAI HumanEval HumanEval/25. Do not edit by hand.
from solution import *  # noqa: F401,F403 (prompt helpers may be referenced by tests)
from solution import factorize as candidate

METADATA = {
    'author': 'jt',
    'dataset': 'test'
}


def check(candidate):
    assert candidate(2) == [2]
    assert candidate(4) == [2, 2]
    assert candidate(8) == [2, 2, 2]
    assert candidate(3 * 19) == [3, 19]
    assert candidate(3 * 19 * 3 * 19) == [3, 3, 19, 19]
    assert candidate(3 * 19 * 3 * 19 * 3 * 19) == [3, 3, 3, 19, 19, 19]
    assert candidate(3 * 19 * 19 * 19) == [3, 19, 19, 19]
    assert candidate(3 * 2 * 3) == [2, 3, 3]


def test_he_025():
    check(candidate)

You are an expert programmer. Solve the task exactly as specified. Output your solution as fenced code blocks using the required file name(s) and the exact function/type signatures requested. Prefer correctness; do not include prose outside code unless asked.

```python
from typing import List


def factorize(n: int) -> List[int]:
    """ Return list of prime factors of given integer in the order from smallest to largest.
    Each of the factors should be listed number of times corresponding to how many times it appeares in factorization.
    Input number should be equal to the product of all factors
    >>> factorize(8)
    [2, 2, 2]
    >>> factorize(25)
    [5, 5]
    >>> factorize(70)
    [2, 5, 7]
    """
    factors = []
    factor = 2
    while factor * factor <= n:
        while n % factor == 0:
            factors.append(factor)
            n //= factor
        factor += 1
    if n > 1:
        factors.append(n)
    return factors
```

The user wants me to implement a function `factorize(n: int) -> List[int]` that returns the prime factorization of a given integer `n` as a list of prime factors in ascending order, with each factor appearing as many times as it divides `n`.

Let me think through the algorithm:

1. Start with the smallest prime (2)
2. While `n` is divisible by the current factor, add it to the result list and divide `n` by that factor
3. Move to the next potential factor
4. Continue until `n` becomes 1

Here's my approach:
- Handle edge cases: if n <= 1, return empty list (though the problem seems to assume positive integers > 1)
- Start with factor = 2
- While factor * factor <= n:
  - While n % factor == 0:
    - Add factor to result
    - n //= factor
  - factor += 1
- If n > 1 after the loop, add n to result (it's a prime)

Let me trace through the examples:
- factorize(8): factor=2, 8%2==0 → [2], n=4; 4%2==0 → [2,2], n=2; 2%2==0 → [2,2,2], n=1; factor=3, 3*3>1, exit; n=1, don't add. Result: [2,2,2] ✓
- factorize(25): factor=2, 25%2!=0; factor=3, 25%3!=0; factor=4, 25%4!=0; factor=5, 5*5<=25, 25%5==0 → [5], n=5; 5%5==0 → [5,5], n=1; factor=6, 6*6>1, exit. Result: [5,5] ✓
- factorize(70): factor=2, 70%2==0 → [2], n=35; factor=3, 35%3!=0; factor=4, 35%4!=0; factor=5, 5*5<=35, 35%5==0 → [2,5], n=7; factor=6, 6*6>7, exit; n=7>1, add → [2,5,7]. Result: [2,5,7] ✓

This looks correct. Let me write the solution.