doesnt work for me :/

Amazing solution, I didn't think of simply replacing the unsorted odd values with the sorted ones.

@MuaazK - that sounds like a good idea. I forked a solution using your suggestion.

If the array's length is 10,000,000 and there are only 5 odd numbers in the first 10 positions, the 'for' loop will put them back in their places and continue to scan the array for no reason.
you should check: "s < sortedOdd.length" and break when it is false.

I did the same

• means more than one however ? sign means if more ok if not ok doesnt matter

The bound + 1 / 2 is a nice trick that is useful in many situations where you need the same index for odd and even lengths.

Nice one.
It took me 3 days to solve this one and i came up with 78 lines of code and then i get here and see it solved in 3 lines :))
Still feeling proud to have solved it :)