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.

To solve this problem, we need to implement a function that returns the prime factors of a given integer in ascending order. Each prime factor should appear as many times as it divides the given integer.

### Approach
The approach to solve this problem involves the following steps:

1. **Initialization**: Start with an empty list to store the prime factors.
2. **Iterate through possible divisors**: Begin with the smallest prime number, which is 2, and check if it divides the given integer.
3. **Divide and Collect Factors**: For each divisor, while the given integer is divisible by the divisor, append the divisor to the factors list and divide the integer by this divisor.
4. **Increment Divisor**: Move to the next possible divisor once the current divisor no longer divides the integer.
5. **Handle Remaining Prime**: After processing all possible divisors up to the square root of the integer, if the remaining integer is greater than 1, it is a prime factor and should be added to the list.

This method ensures that we efficiently find all prime factors by iterating through possible divisors and reducing the problem size at each step.

### Solution Code

```python
from typing import List

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

### Explanation
- **Initialization**: We start with an empty list `factors` to store the prime factors.
- **Divisor Loop**: We use a while loop to iterate through possible divisors starting from 2. The loop continues as long as the square of the current divisor is less than or equal to the given integer.
- **Division Check**: For each divisor, we check if it divides the integer. If it does, we append the divisor to the factors list and divide the integer by this divisor repeatedly until it no longer divides evenly.
- **Increment Divisor**: Once the current divisor no longer divides the integer, we move to the next possible divisor by incrementing the divisor.
- **Remaining Prime Check**: After processing all possible divisors up to the square root of the integer, if the remaining integer is greater than 1, it is a prime factor and is added to the factors list.

This approach efficiently finds all prime factors by reducing the problem size at each step and ensures that each factor is listed the correct number of times.

Okay, I need to write a Python function called factorize that takes an integer n and returns a list of its prime factors in order from smallest to largest. Each factor should appear as many times as it divides n. So, for example, factorize(8) should return [2,2,2], and factorize(25) returns [5,5].

Hmm, how do I approach this? Well, I remember that prime factorization involves dividing the number by the smallest possible primes until we're left with 1.

Let me think about the steps. I'll start with the smallest prime, which is 2. I'll divide n by 2 as many times as possible, adding 2 to the factors list each time. Then move to the next possible divisor, which would be 3, then 5, and so on.

Wait, but how do I efficiently find the next divisor? Maybe I can loop through possible divisors starting from 2, and for each, while the current number is divisible by that divisor, append it to the factors and divide n by it. Once it's no longer divisible, move to the next possible divisor.

But wait, after 2, the next possible divisor is 3, then 5, but wait, what about 4? Well, since 4 is 2 squared, if n is divisible by 4, it must have been divided by 2 already until it's no longer possible. So after 2, the next possible divisor is 3, then 5, etc.

So the algorithm could be:

Initialize an empty list for factors.
Start with the smallest prime, 2.
While i * i <= n:
    while n is divisible by i, append i to factors and divide n by i.
    increment i by 1.
If n is still greater than 1 after this loop, append it to factors.

Wait, but that might not cover all cases. Let me test this logic.

Take n=70. The factors should be [2,5,7].

Let's see:
i starts at 2. 70 is divisible by 2, so add 2, n becomes 35.
Check 35 divisible by 2? No. So i increments to 3. 35 not divisible by 3. i becomes 4. 35 not divisible by 4. i becomes 5. 35 is divisible by 5, add 5, n becomes 7. Now, 7 is not divisible by 5. i increments to 6, but 6^2 is 36 which is larger than 7. So loop ends. Then, since n is 7>1, add 7. So factors are [2,5,7], which is correct.

Another test case: n=8. i=2, 8/2=4, add 2. 4/2=2, add 2. 2/2=1, add 2. Now n is 1, so loop ends. So factors are [2,2,2], correct.

Another test: n=25. i=2, 25 not divisible by 2. i=3, not divisible. i=4, 25 not divisible by 4. i=5, 25/5=5, add 5. Then 5/5=1, add 5. So factors [5,5], correct.

What about n=1? Well, according to the problem statement, the input is an integer, but the examples start from 8. Wait, the function is supposed to handle any integer, but what if n is 1? The function should return an empty list because 1 has no prime factors. Let me see: in the code, after the loop, n would be 1, so the condition n>1 is false, so nothing is added. So factors would be empty, which is correct.

Another test case: n=12. Factors should be [2,2,3]. Let's see: i=2, 12/2=6, add 2. 6/2=3, add 2. Now i increments to 3. 3 is <= sqrt(3) (which is ~1.732). Wait, 3^2 is 9, which is larger than 3. So the loop ends, and n is 3>1, so add 3. So factors are [2,2,3], correct.

Wait, but in the loop condition, it's while i*i <=n. So for i=3, n is 3. i*i is 9, which is larger than 3, so the loop doesn't run. Then, since n is 3>1, it's added. So yes.

So the algorithm seems solid.

Now, how to implement this in Python.

The function will start with an empty list. Then, i starts at 2. While i*i <=n, loop:

Wait, but wait: what if n is reduced to a value smaller than i*i during the loop? For example, when n is 7, and i is 3. 3^2 is 9>7, so the loop stops, and 7 is added.

So the code outline is:

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

Wait, but wait: what about when n is 0 or 1? Let's see.

If n is 0, the function will loop indefinitely because i starts at 2, and 2*2 is 4>0, so the loop doesn't run. Then, since n is 0>1 is false, nothing is added. So factors is empty. But 0 can't be factorized into primes. So perhaps the function should handle n=0 as a special case. But the problem statement says the input is an integer, but the examples are positive. So perhaps the function is intended for n >=2.

But the problem statement says, "Input number should be equal to the product of all factors." So for n=1, the product is 1, which is correct. But for n=0, the product is 0, but the factors list would be empty, which is incorrect. So perhaps the function should handle n=0 by returning an empty list, but the problem statement may not require that.

But the problem statement's examples don't include 0 or 1. Let's see the sample inputs:

Sample 1: 8 → [2,2,2], product is 8.

Sample 2:25 → [5,5], product 25.

Sample3:70 → [2,5,7], product 70.

So the function is intended for n >=2.

But what about n=1? The function returns an empty list, which is correct because 1 has no prime factors.

So the code as written should handle all cases correctly.

Testing the code:

Let me test n=12. The code should return [2,2,3].

Yes, as discussed earlier.

Another test: n=17, which is prime. The loop runs i=2: 2^2=4 <=17. 17%2 is 1, so no division. i increments to 3. 3^2=9 <=17. 17%3 is 2, no division. i=4: 16 <=17. 17%4 is 1. i=5: 25>17, loop ends. Then n=17>1, so added. So factors [17], correct.

What about n=6: factors [2,3], correct.

So the code seems correct.

Now, implement this in the function.

Wait, but in Python, for n=0, the loop will not run, and then n>1 is false, so factors is empty. So 0 returns empty list, which is incorrect because 0 has infinitely many factors. But perhaps the function is intended for n >=2, so this is acceptable.

So the function is as written.