Nice!

Is that solution more efficient than going for each character and checking? The ToUpper method doesn't need to do a for each too? The best scenario of checking is O(1), case the first character is lower, it returns, if ToUpper does use a for than it would be O(N) always. What do you think?

Thanks!

Brilliant!

Nice divmod usage!

Clever!!

Nice

absouloutly clever, I never knew about divmod, I have to keep an eye on it, It has so many use cases.

Nice one!

I'm an idiot -_-

this is big brain

oooh.. thats really good. I had to google every thing is this code to understand. Hope I can remember it in 20 minutes :D

The if odd part is an example of Truthy values in Python - you can learn more here:

https://www.freecodecamp.org/news/truthy-and-falsy-values-in-python/

(Not an endorsement of freecodecamp, just the first google search result that looked OK).

In this case, it is a shorter version of if odd == 1, which in turn is determined by the remainder calculation in the step above.

You can also experiment with this in your IDE: for example try:

a = [1,4,39]
b = []

if a:
  print("array a is not empty")
if b:
  print("array b is not empty")

You should find that only the first statement will print.

divmod has an amazing user experience level!

Thanks for the comment. I found the following link which helped me https://docs.python.org/2/library/functions.html#divmod