```>The bubblesort you did here will not be faster, no matter what order
>you traverse the items in. After all, you will look at each item in the
>outer loop and again at each item in the inner loop again.
>That makes it n-squared no matter what order you traverse them in.

Just think. If you go from biggest to smallest in inner loop then
if you put smaller number at the beginning then bigger number
goes somewhere in the list. If you find even smaller number then exchange
repeats but the two numbers you exchanged are in better place, the smaller
number
stands before bigger.
That means that there will be less exchanges needed if you traverse inner
loop
from last to first. Reversing elements will not help.
So it's true that there will be same number of times going through loops but
there will be less exchanges needed so that is why it would be faster.

>[1,2,3,4].reverse.each_with_index

As I said that will not help.
But there is a way to get that speed increment.
The list has to be sorted from biggest to smallest and then do reverse, but
then, reverse takes some time.

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