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Start your free trialBrian S.
3,176 PointsLittle help to pass please
I'm sure i am misunderstanding here.... but, if 'treehouse' (9 letters, odd) should be 'treeHOUSE', then it's implied that given an odd number of letters, more should be upper than lower. Following these rules, when I input my code, I get an error that says 'kenneth' (7 letters, odd) should be 'kennETH', implying that given an odd number of letters, more should be lower. Any suggestions, or clarifications? Thank you.
def sillycase(word):
half = len(word) // 2
if len(word) % 2 != 0: #its odd
return word[0:half].lower() + word[half:].upper()
else: # its even
half = len(word) // 2
return word[0:half].lower() + word[half:].upper()
print('even')# The first half of the string, rounded with round(), should be lowercased.
# The second half should be uppercased.
# E.g. "Treehouse" should come back as "treeHOUSE"
1 Answer
Steven Parker
231,275 PointsThe challenge asks for the "first half of the string, rounded with round()..."
But you calculated your half using the floored division operator ("//
").
If you calculate the half size as suggested, you should pass the challenge. Also, you won't need to make a special case for even vs. odd.