I was doing some basic problems to brush up my programming skills(its been 2 years). I was solving the question when an alternate method struck me.

1) When you rotate an array by, say 3 elements, the last n — 3 elements are directly moved up to the front. Keeping this in mind, I swapped the elements at 3 index difference.

2) This led to the last 3 elements not being in correct order, and actually offset by n % 3. So applying the same logic to them that was applied to the whole array, and repeating.

3) There will be exactly n — 1 operations (initial n — 3, then 1 (+ 1) to rearrange the last 3 elements), thus it is O(n), and memory of O(1).

static void rotateArr(int[] arr, int d) {
    int n = arr.length, j = 0, h = n, z = n;
    if (d % n != 0) {
        d %= n;
        while (d != n) {
           for (int i = j; i < n - d + j; i++) {
                int temp = arr[i];
                arr[i] = arr[i + d];
                arr[i + d] = temp;
            }
            j = z - d;
            n = d;
            d = d - h % d;
            h = n;
        }
    }
}