how do you get to the amount of the combinations?
for l=4 there are more than 4 combinations:
(((xx)x)x)
((xx)(xx))
((x(xx))x)
(x((xx)x))
(x(x(xx)))

But the input is always the "lorem ipsu" text isnt it? Onlz the width changes.Im not sure what the test cases test.

you can print the input the very same way you print your solution, to see which cases your code is failing

I pass 684 tests and fail 15. What do the test cases test? Or how can i see the expected output to know whats wrong? When I print my solution it looks perfectly aligned and justified. Im using Python if thats relevant. Thanks!

Can someone explain to me how there can be more than 300 possible parenthesis combinations for an "s" of length 8?

There should be at most 64 combinations for "s" of length 8, and some of them will be false.

If you go by the examples provided in the description everything must be enclosed in parenthesis. And the number of combinations of parenthesis go 1,2,4,8,16, etc for length 2,3,4,5,6, etc. It doesnt seem to make sense.

What am I missing? Thanks.

Im exactly in the same boat as you, did you ever figure that out? If you go by the examples provided in the intro everything must be enclosed in parenthesis. And the combinations of parenthesis go 1,2,4,8,16, etc for length 2,3,4,5,6, etc. It desnt seem to make sense.

Nice solution. Cant there be more than 2 most common weekdays? Thanks.

Good guy Gerg

If you already returned something, your function is already "finished", so you will never get to the final (unconditional) return.

Question: Wont you always get a "Lost" returned in addition to a "Dead" or "Finish"? Since whatever happens you have an unconditional "return Lost" at the end. Thanks!