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]:
    factors = []
    d = 2
    while d * d <= n:
        while n % d == 0:
            factors.append(d)
            n //= d
        d += 1
    if n > 1:
        factors.append(n)
    return factors
```

The user wants a Python function `factorize(n: int) -> List[int]` that returns the prime factorization of `n` as a list of prime factors in non-decreasing order.
The product of the factors should equal `n`.
The input `n` is an integer. I should assume `n >= 2` based on typical HumanEval problems, but I'll handle `n <= 1` gracefully if needed, though usually `n >= 2`.

Algorithm:
1. Initialize an empty list `factors`.
2. Start with divisor `d = 2`.
3. While `d * d <= n`:
   While `n % d == 0`:
     Append `d` to `factors`.
     `n //= d`
   `d += 1`
4. If `n > 1`, append `n` to `factors`.
5. Return `factors`.

Let's test with examples:
`factorize(8)`:
d=2, 8%2==0 -> factors=[2], n=4
4%2==0 -> factors=[2,2], n=2
2%2==0 -> factors=[2,2,2], n=1
d=3, 3*3 > 1, loop ends.
n=1, not > 1.
Returns [2, 2, 2]. Correct.

`factorize(25)`:
d=2, 25%2!=0, d=3
d=3, 25%3!=0, d=4
d=4, 25%4!=0, d=5
d=5, 5*5 <= 25. 25%5==0 -> factors=[5], n=5
5%5==0 -> factors=[5,5], n=1
d=6, 6*6 > 1, loop ends.
Returns [5, 5]. Correct.

`factorize(70)`:
d=2, 70%2==0 -> factors=[2], n=35
d=3, 35%3!=0, d=4
d=4, 35%4!=0, d=5
d=5, 5*5 <= 35. 35%5==0 -> factors=[2,5], n=7
d=6, 6*6 > 7, loop ends.
n=7 > 1 -> factors=[2,5,7]
Returns [2, 5, 7]. Correct.

Edge cases:
`n=1`: returns `[]`. Product of empty list is 1. Usually acceptable.
`n` is prime: e.g., 7. d=2,3. Loop ends. n=7>1 -> returns [7]. Correct.

Complexity: O(sqrt(n)) which is fine for typical constraints.

One minor optimization: handle 2 separately, then iterate odd numbers. Not strictly necessary but good practice. I'll stick to the simple version for clarity unless performance is critical, but HumanEval usually has small inputs. The simple version is fine.

Let's write out the code.

```python
from typing import List

def factorize(n: int) -> List[int]:
    factors = []
    d = 2
    while d * d <= n:
        while n % d == 0:
            factors.append(d)
            n //= d
        d += 1
    if n > 1:
        factors.append(n)
    return factors
```

Check constraints/types: `n: int`, returns `List[int]`. Matches signature.
The docstring examples match.
I will wrap it in the requested format.

One thing: HumanEval problems sometimes have `n` up to large values, but `d += 1` is fine. If `n` is large, iterating only odds after 2 is better. I'll stick to `d += 1` for simplicity, it's standard. Actually, `d += 1` is perfectly fine for Python in these contexts.

Wait, should I consider negative numbers or 0? The problem says "prime factors of given integer", usually implies positive integers > 1. I'll assume `n >= 2`. If `n <= 1`, returning `[]` is standard.

Let's double check logic.
`factorize(8)` -> `[2, 2, 2]`
`factorize(25)` -> `[5, 5]`
`factorize(70)` -> `[2, 5, 7]`
All good.

I will produce the final code.
No extra fluff.
Code matches exactly.
Ready. 
One minor point: `from typing import List` is already in the prompt, but I'll include it in the solution file as requested. The prompt says "Implement a file solution.py that completes the function below." I will provide the full code.

Final check of the code:
```python
from typing import List

def factorize(n: int) -> List[int]:
    factors = []
    d = 2
    while d * d <= n:
        while n % d == 0:
            factors.append(d)
            n //= d
        d += 1
    if n > 1:
        factors.append(n)
    return factors
```
This is correct and robust.
I will output this. 
All steps verified.
Proceeds. 
Output matches request.✅
Note: I'll make sure indentation and formatting are correct.
The prompt asks to keep the given name and signature.
I'll output exactly the code block.
Done. 
Self-Correction/Verification during thought: 
Is there any case where `d` increments slowly and causes TLE? For `n` up to 10^9, `sqrt(n)` is ~31622, which is trivial. Even for 10^12, it's 10^6, which is fine in Python. HumanEval inputs are usually small. So this is optimal enough.
I'll stick with it.
Output matches. 
Proceeds. 
[Output Generation] -> *Proceeds*
```python
from typing import List

def factorize(n: int) -> List[int]:
    factors = []
    d = 2
    while d * d <= n:
        while n % d == 0:
            factors.append(d)
            n //= d
        d += 1
    if n > 1:
        factors.append(n)
    return factors
```
Looks good.✅
Done. 
Output matches response. 
(Self-Correction/Refinement during generation prep)
Wait, should I handle `n=1`? The problem says "Input number should be equal to the product of all factors". For `n=1`, product of `[]` is 1 by convention, but sometimes they expect `[]`. It's fine.
The code is standard.
I will produce it. 
All good. 
Output Generation. 
[Done]