Find the minimum of a sorted and shifted array with better than O(n) time complexity

Question

We have an assignment to search for the minimum element of a sorted array that is shifted to the right afterwards. For example: [1, 5, 6, 19, 56, 101] becomes [19, 56, 101, 1, 5, 6]. The method should be implemented using a divide and conquer algorithm and it should have a better asymptotic time complexity than O(n). EDIT: I forgot to add that the elements int the array are unique.

I already implemented a method and wanted to ask if this is better than O(n) and if there are ways to improve my method.

public class FindMinimum {
    public void findMinimum(int[] arr) {

        // the recursive method ends when the length of the array is smaller than 2
        if (arr.length < 2) {
            return;
        }

        int mid = arr.length / 2;

        /*
         * if the array length is greater or the same as two, check if the middle
         * element is smaller as the element before that. And print the middle element
         * if it's true.
         */
        if (arr.length >= 2) {
            if (arr[mid - 1] > arr[mid]) {
                System.out.println("Minimum: " + arr[mid]);
                return;
            }
        }

        /*
         * separate the array in two sub-arrays through the middle and start the method
         * with those two arrays again.
         */
        int[] leftArr = new int[mid];
        int[] rightArr = new int[arr.length - mid];
        for (int i = 0; i < mid; i++) {
            leftArr[i] = arr[i];
        }
        for (int i = mid; i < arr.length; i++) {
            rightArr[i - mid] = arr[i];
        }
        findMinimum(leftArr);
        findMinimum(rightArr);
    }
}

Answer 1

In Java you could use a List because than you can create a Sublist.

private Integer findMinimum(List<Integer> list) {
    if (list.size() < 2)
        return list.get(0);

    int mid = list.size() / 2;

    // create left and right list
    List<Integer> leftList = list.subList(0, mid);
    List<Integer> rightList = list.subList(mid, list.size());

    if (leftList.get(leftList.size() - 1) <= rightList.get(rightList.size() - 1))
        return findMin(leftList);
    else
        return findMin(rightList);
}

When you create a Sublist with Java there is no copy. So to create a new Sublist takes a complexity of O(1). So the function has a complexity of O(logn).