better yet, u.equals(str)

While I wouldn't do the check using root of n but instead the square of the running index since that is faster, he is generally right to just check until the squareroot since every divisor you find already has another factor on the other side of the root you need to multiply with to get n. Think of this as an area of same sice but different side length were you shorten one side so the other get's longer. When you arrive at the perfect square, you don't have to go further since all other sidelengths were already checked "the other way round" (hence add 2 to the output for every divisor except for the root - there the other factor is identical).

If you build up a string like this with repeated "+", I would highly recommend to use StringBuilder, since otherwise you store binaryN.Length different strings.

funny

Maybe you used percentile_disc instead of percentile_cont?

write "string" with lowercase "s"

It would be better to use StringBuilder rather than adding strings, since strings ar immutable i.e. for every loop step a new string must be stored.

the Link was updated to https://blog.ostermiller.org/finding-a-loop-in-a-singly-linked-list/

the Link was updated to https://blog.ostermiller.org/finding-a-loop-in-a-singly-linked-list/

and now for a shift of 12 letters :)

Made my day

You're absolutely right, there was a mistake in my code. It should be fixed now. Please try again, and thanks for notification!!

In ruby it makes you unshift for numbers.