"""
Response cleaner module.
Removes markdown formatting and special characters from AI responses.
"""
import re


def clean_response(text: str) -> str:
    """
    Clean AI response from markdown formatting and special characters.
    
    Args:
        text: Raw AI response with possible markdown
        
    Returns:
        Clean text suitable for TTS
    """
    if not text:
        return ""
    
    # Remove citation references like [1], [2], [citation], etc.
    text = re.sub(r'\[\d+\]', '', text)
    text = re.sub(r'\[citation\s*needed\]', '', text, flags=re.IGNORECASE)
    text = re.sub(r'\[source\]', '', text, flags=re.IGNORECASE)
    
    # Remove markdown bold **text** and __text__
    text = re.sub(r'\*\*(.+?)\*\*', r'\1', text)
    text = re.sub(r'__(.+?)__', r'\1', text)
    
    # Remove markdown italic *text* and _text_
    text = re.sub(r'\*(.+?)\*', r'\1', text)
    text = re.sub(r'(?<!\w)_(.+?)_(?!\w)', r'\1', text)
    
    # Remove markdown strikethrough ~~text~~
    text = re.sub(r'~~(.+?)~~', r'\1', text)
    
    # Remove markdown headers # ## ### etc.
    text = re.sub(r'^#{1,6}\s*', '', text, flags=re.MULTILINE)
    
    # Remove markdown links [text](url) -> text
    text = re.sub(r'\[([^\]]+)\]\([^)]+\)', r'\1', text)
    
    # Remove markdown images ![alt](url)
    text = re.sub(r'!\[([^\]]*)\]\([^)]+\)', '', text)
    
    # Remove inline code `code`
    text = re.sub(r'`([^`]+)`', r'\1', text)
    
    # Remove code blocks ```code```
    text = re.sub(r'```[\s\S]*?```', '', text)
    
    # Remove markdown list markers (-, *, +, numbered)
    text = re.sub(r'^\s*[-*+]\s+', '', text, flags=re.MULTILINE)
    text = re.sub(r'^\s*\d+\.\s+', '', text, flags=re.MULTILINE)
    
    # Remove blockquotes
    text = re.sub(r'^\s*>\s*', '', text, flags=re.MULTILINE)
    
    # Remove horizontal rules
    text = re.sub(r'^[-*_]{3,}\s*$', '', text, flags=re.MULTILINE)
    
    # Remove HTML tags if any
    text = re.sub(r'<[^>]+>', '', text)
    
    # Remove extra whitespace
    text = re.sub(r'\n{3,}', '\n\n', text)
    text = re.sub(r'  +', ' ', text)
    
    # Clean up and return
    text = text.strip()
    
    return text