he-001

# HumanEval/1

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 separate_paren_groups(paren_string: str) -> List[str]:
    """ Input to this function is a string containing multiple groups of nested parentheses. Your goal is to
    separate those group into separate strings and return the list of those.
    Separate groups are balanced (each open brace is properly closed) and not nested within each other
    Ignore any spaces in the input string.
    >>> separate_paren_groups('( ) (( )) (( )( ))')
    ['()', '(())', '(()())']
    """
```

<!-- imported from OpenAI HumanEval (HumanEval/1) -->

# Auto-generated from OpenAI HumanEval HumanEval/1. Do not edit by hand.
from solution import *  # noqa: F401,F403 (prompt helpers may be referenced by tests)
from solution import separate_paren_groups as candidate

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


def check(candidate):
    assert candidate('(()()) ((())) () ((())()())') == [
        '(()())', '((()))', '()', '((())()())'
    ]
    assert candidate('() (()) ((())) (((())))') == [
        '()', '(())', '((()))', '(((())))'
    ]
    assert candidate('(()(())((())))') == [
        '(()(())((())))'
    ]
    assert candidate('( ) (( )) (( )( ))') == ['()', '(())', '(()())']


def test_he_001():
    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 separate_paren_groups(paren_string: str) -> List[str]:
    paren_string = paren_string.replace(" ", "")
    result = []
    current = []
    balance = 0
    for char in paren_string:
        if char == '(':
            current.append(char)
            balance += 1
        elif char == ')':
            current.append(char)
            balance -= 1
            if balance == 0:
                result.append(''.join(current))
                current = []
    return result
```