Rotate a circular price buffer by k

Question

A ring buffer of the last n prices needs to be rotated right by k positions so the newest slot lines up at the front. Do it in place with constant extra space. Rotate the array right by k in place. Aim for O(n) time and O(1) space.

Accepted Answer

Rotating right by k moves the last k elements to the front. The reversal trick does this without a temp array: reverse the whole array, then reverse the first k elements and the remaining n - k elements separately. Reduce k modulo n first so large rotations wrap correctly. Tracing [1, 2, 3, 4, 5] with k = 2: reversing gives [5, 4, 3, 2, 1]; reversing the first two gives [4, 5, 3, 2, 1]; reversing the rest gives [4, 5, 1, 2, 3]. Time O(n), space O(1) (the slice reversals swap in place). Edge cases: an empty array is returned unchanged, and taking k %= n prevents index errors when k = n or k is…

Rotate a circular price buffer by k

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