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# Discussion on: Daily Challenge #124 - Middle Me

Gabriel Romualdo • Edited

My solution in Python:

def middle_me(x, y, n):
if(n % 2 != 0):
return x
else:
return ( y * (n / 2) ) + x + ( y * (n / 2) )

Which can be abbreviated to:

def middle_me(x, y, n):
return x if n % 2 != 0 else ( y * (n / 2) ) + x + ( y * (n / 2) )

Hoped you liked this solution, definitely not the most efficient or clean, but I have been working on my Python skills as I am mainly a web developer working with JS and ES6.

— Gabriel

Rafael Acioly

Nice! just a hint, if you have a return inside the first if you don't need the else ;)

Here's my solution:

def middle_me(middle: str, repeat: str, repeat_quantity: int) -> str:
if repeat_quantity % 2 != 0:
return middle

half = repeat_quantity // 2
content = repeat * repeat_quantity
return content[:half] + middle + content[half:]
Michael Kohl

The string slicing seems to be extra effort, how about something like this?

padding = repeat * (repeat_quantity // 2)