Make sure that there is a whitespace. It's "' '{3}" not "{3}". :)

doesnt work for me :/

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

3 spaces you could use " {3}" instead~

@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

it checks whether there are more than one whitespace and splits

StringBuffer is way slower than StringBuilder, because it is synchronized. And if nowadays you see StringBuffer anywhere in single threaded code it is archaic (or poorly written) code.
BTW code from kata with +="" whould compile to StringBuilder concentrations, very inefficient ones however.

first ,U should add this code:

morseCode.isEmpty() return "";

second ,
StringBuffer result
result.append()

how this works? word.split("\s+")?

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 :)