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]:
    """ 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('( ) (( )) (( )( ))')
    ['()', '(())', '(()())']
    """
    # Remove all spaces from the input string
    paren_string = paren_string.replace(' ', '')
    
    result = []
    current_group = []
    balance = 0
    
    for char in paren_string:
        if char == '(':
            balance += 1
            current_group.append(char)
        elif char == ')':
            balance -= 1
            current_group.append(char)
            
            # When balance returns to 0, we've completed a group
            if balance == 0:
                result.append(''.join(current_group))
                current_group = []
    
    return result
```

Let me understand the problem:

1. We have a string containing multiple groups of nested parentheses
2. We need to separate these groups into individual strings
3. Each group is balanced (each open parenthesis has a matching close parenthesis)
4. Groups are not nested within each other (they're at the same level)
5. We should ignore spaces in the input string

For example: `( ) (( )) (( )( ))` should become `['()', '(())', '(()())']`

The approach:
1. Remove all spaces from the input string
2. Iterate through the string, tracking the balance of parentheses
3. When balance goes from 0 to 1, we start a new group
4. When balance goes from 1 to 0, we end a group
5. Collect each group when it ends

Let me trace through the example:
- Input: `( ) (( )) (( )( ))`
- After removing spaces: `()(( ))(( )( ))` → `()(( ))(( )( ))` → `()(( ))(( )( ))`
- Actually: `()(( ))(( )( ))` → `()(( ))(( )( ))`

Wait, let me be more careful:
- Original: `( ) (( )) (( )( ))`
- Remove spaces: `()(( ))(( )( ))` → `()(( ))(( )( ))`

Actually: `( ) (( )) (( )( ))`
Remove spaces: `()(( ))(( )( ))`

Let me count:
- Position 0: `(` → balance = 1, start group
- Position 1: `)` → balance = 0, end group, save `()`
- Position 2: `(` → balance = 1, start group
- Position 3: `(` → balance = 2
- Position 4: `)` → balance = 1
- Position 5: `)` → balance = 0, end group, save `(())`
- Position 6: `(` → balance = 1, start group


- Position 7: `(` → balance = 2
- Position 8: `)` → balance = 1
- Position 9: `(` → balance = 2
- Position 10: `)` → balance = 1
- Position 11: `)` → balance = 0, end group, save `(()())`

I'll track the balance carefully, noting when it transitions from 0 to 1 (starting a group) and from 1 to 0 (ending a group). This helps me identify each balanced parenthetical group.

The key is maintaining a running balance counter and capturing substrings when the balance returns to zero.