Discussion on: Daily Challenge #124 - Middle Me

[Gabe Romualdo]

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:]