Looks good to me. But you haven't shown how you update pb so I can't understand completely.

One note, it seems to me that pb=(b>pb)?b:pb; is another way to write pb=max(b,pb);. I note that using algebra, pb+c>c is equivalent to pb>0, which is slightly more efficient. Similarly pb+c>pb is the same as c>0.

On the other hand, when is is not the best sum. you will need to update pb. When there are all negative elements, pb is undefined by the problem statement; either pb = 0 because the shortest array is 0 length, or it's equal to the smallest element, because the shortest array is 1 length.

I think that you used the second definition, right?

For noncontiguous with at least one positive element, similar to @maximshen's solution above, just needs to sum all positive elements. That one is easy, it looks like: for(int x:arr) { b = b + (c>0)?c:0; }