Can write a c++ solution as an ECT practice. Easy to verify. Added to enroute.txt.

[[EPI300]] 6.13 is probably a much shorter and more readable solution!

A: find the largest common denominator between N and K. If it’s 3, then the array consists of 3 interleaved subsequences (abbreviations). We will “process” each subsequences simultaneously since they are completely independent. First subsequences consists of Slots 0,3,6,9… Within each subsequences , K’ and N’ (scaled down by the denominator) have no common denominator, therefore an iteration will visit every element exactly once until we revisit the first element.

Q: rotate a large array (N elements) by K positions. Both N and K are potentially large, so we need 1) time efficiency 2) memory efficiency. Ideally, O(1) space and O(N) time.

–analysis

If K = 1, then just shift every element right, rotating the last element to the head. If K = 2, just repeat once. But K is large.

You can assume it’s an array of pointers, or you can assume it’s an array of integers — Same challenge.

The brute-force solution requires O(K*N), shifting all N elements K times.

Without loss of generality, let’s rotate by 5 slots. If N is multiple of 5, like 30, then there are 5 individual, independent, interleaved arrays of 10. Just rotate each by 1 slot.

If N is not a multiple of 5, like 31, then view the array as a circular array and we hop by 5 slots each time. So our journey will start from position 0 (or any position) and cover every position and come back to the starting position. This can be proven by contradiction —

Sooner or later we must be re-visiting some position. We know that position is reachable from position 0, so the distance between them must be multiple of 5. So position 0 must be revisited.