nice approach

https://docs.oracle.com/javase/8/docs/api/java/util/List.html#toArray-T:A-

Why is 'new String[0]' passed in as parameter to the toArray method. What does it even mean? Thanks in advance.

Because at that point index i+1 is same as string length.

And exception is not thrown by substring unless index is greater than string length

Refer: https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#substring(int)

Can someone explain how when the i gets to the last letter (str.length -1) there is not an out of bounds exception caused by the str.substring(i+1) ?

Compare:

1. "if ch is blank then skip", VS.
2. "if ch is not blank then do something"

Personally, I find option 1 is more natural way to describe the alogrithm. YMMV.

Is the use of the keyword "continue" better than using brackets in this case?

Thank you, after correctly copied the array the kata showed no errors.

Do a console.log of numbers before the return line, you've mutated the input. var arrayCopy = numbers; //Copy the array, just in case doesn't make a copy, both vars point to the same array. Try looking in the web how to make a swallow copy of an array.

Or you could try creating a new array instead, or both ;)

Post your code, mark your post as having spoiler content.

I'm having troubles with this, the random test cannot be passed because it tooks the 2nd and 3rd element from the array, and I didn't changed the original array, here are some samples from the random test

Log
[ 1038, 40, 62, 55, 38 ]
78
Expected: 117, instead got: 78
Log
[ 1000, 27, 32, 25, 40 ]
52
Expected: 72, instead got: 52
Log
[ 1067, 65, 92, 50, 96 ]
115
Expected: 188, instead got: 115
Log
[ 1087, 73, 32, 31, 100 ]
63
Expected: 173, instead got: 63
Log
[ 1044, 72, 22, 19, 21 ]
40
Expected: 94, instead got: 40
Log
[ 1076, 26, 80, 97, 0 ]
26
Expected: 177, instead got: 26