@yong.cao.7731 Yours was a great question. Here is how I understood this part.

Take the example

A-3, B-3,C-3,D-2 (counts for each task)

and n=2 (cooling period).

Now the arrangement for A,B,C is:

A B C A B C A B C

they can be tightly arranged with no gaps using the cooling period effectively to schedule other tasks.

What @beetlecamera is trying to explain is, n=2 is the minimum cooling period for two same tasks to execute. We can always insert more than 2 tasks also. So, then we could do this:

A B C D A B C D A B C

As you can see now, we have three tasks between two successive executions of A making the total slots needed equal to array size. (Instead of A B C A B C A B C D _ _ D which would need 13 slots).

Hope this helps! cheers