the step values fixed?
this "1" ?binary search
in your hands. you can predict the value of any
as the original array, and fragment. the deviation is possible to conclude that there is a gap (or several)
we must remember that Dvina the sample will be effective in determination of continuous fragments, further breaks may require a modification of the heuristic algorithm
ps for sure!.. in General, all the arguments about arithmetic progressions and binary search is true for any step of the progression (not just 1)