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]
    """
    if n <= 1:
        return []
    
    factors = []
    divisor = 2
    
    while n > 1:
        while n % divisor == 0:
            factors.append(divisor)
            n //= divisor
        divisor += 1
        
        # Optimization: if divisor * divisor > n, then remaining n is prime
        if divisor * divisor > n:
            if n > 1:
                factors.append(n)
            break
    
    return factors
```

The task is to implement a function `factorize(n: int) -> List[int]` that returns the prime factorization of a given integer `n`. The factors should be returned in order from smallest to largest, with each factor appearing as many times as it divides the number.

For example:
- `factorize(8)` should return `[2, 2, 2]` because 8 = 2 × 2 × 2
- `factorize(25)` should return `[5, 5]` because 25 = 5 × 5
- `factorize(70)` should return `[2, 5, 7]` because 70 = 2 × 5 × 7

The algorithm:
1. Start with the smallest prime (2)
2. While the current factor divides n, add it to the result and divide n by it
3. Move to the next potential factor
4. Continue until n becomes 1

I need to handle edge cases:
- What if n is 1? The prime factorization of 1 is typically an empty list
- What if n is negative? The problem doesn't specify, but typically we'd handle positive integers
- What if n is 0? This would be undefined

Let me write the solution:

```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]
    """
    if n <= 1:
        return []
    
    factors = []
    divisor = 2
    
    while n > 1:
        while n % divisor == 0:
            factors.append(divisor)
            n //= divisor
        divisor += 1
        
        # Optimization: if divisor * divisor > n, then n is prime
        if divisor * divisor > n:
            if n > 1:
                factors.append(n)
            break
    
    return factors
```

I'll optimize the factorization by checking if the divisor squared exceeds the remaining number. This allows me to quickly identify when the remaining number is prime, avoiding unnecessary iterations.

The key insight is that if no divisor less than or equal to the square root of n divides n, then n itself must be prime. This significantly reduces the number of iterations needed to find all prime factors.