You can also skip the "i < (int)strlen(full_name)" check because a pre-condition of the input is that a space must exist.

Nice combinations code.

Avoid recursion if you can.

This is hilarious and clever. Good job man.

Thanks, I learned several things from this solution :) Very clean.

I have a question though. Why do you need to free the copy variable but not variables like the char* p from the last iteration?

Mmmmmkay

This is not beautiful, it is a waste of processing resources. Sorting every word is awful. Just Hash it with O(n). Sorting is O(nlog(n))

It means the value will only be replaced if it is nil or false. (Correct me if I'm wrong please, I'm also new)

#Case 1:
a = 1
a ||= 10
print a #=> 1

#Case 2:
a = nil
a ||= 10
print a #=> 10

a||= 10 is logically equivalent to: a || a = 10

Let's understand that behaviour. When we put together many or's:
a || b || c || d, it will return true immediatly when the first true value is found.
So in the example above, if 'a' is true, then b, c and d WON'T be evaluated by the or operator.

The same happens here: a || a = 10.

if a is false (or nil) it will evaluate (a = 10) which then changes the value of a.
if a is true (or != nil), then (a = 10) WON'T be evaluated at all.

Why is this clever? This is just a library... 0% clever.... But a good practice though. Don't reinvent the wheel.

Wow

I did the same solution. I guess they would rather see some clean code than efficient code.

They shouldn't cancel each other. Only adjacent opposite direction cancel each other. That's why he/she only checks the last element in the stack.

Clever girl

Please, I need someone to explain this.... this is crazy. I first tried to avoid loops but couldn't think of any theorem to rely on.