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If K=2n then the loop will run twice and k will be set to 0. If k is a multiple of n then the array after k rotations will be same as initial array. k=k%n could also do the trick.
if k is a multiple of n then won't the remainder be 0?
Yes. Which means, in effect, that no rotation should be performed upon the array, because rotating each element by k will have each of them end up right back where they started. Using 0 for k will have that effect.