Ooh, that's nice. this is much more readable than mine. return min([(-1 * ltext.count(c), c) for c in string.ascii_lowercase])[1] More

Similar to what I did, so I'm biased. Thumbs up. :) More

Nice! Very similar to what I did. Is there an advantage to using a.lower().replace() instead of a.replace().lower()? More

Terrific solution! I aimed for the same logic, but plowed through it the hard way; your code is so much more elegant. I love the use of map()! More

Thanks for the RE approach; I'm still new to these. More

Nice! You can yank the list and [get it all in one loop](http://www.checkio.org/mission/digits-multiplication/publications/bryukh/python-3/convert-and-iterate/?ordering=most_voted), if you want: result = 1 for digit in str(number): result *= int(digit) if int(digit) else 1 # multiply by 1 More