Binary search in a rotated sorted array

RotatedSortedArray.java

public class RotatedSortedArray {
    public static void main(String[] args) {
        testFind();
    }

    public static void testFind() {
        System.out.println(find(null, 10));// Expect -1
        int[] arr1 = new int[]{};
        System.out.println(find(arr1, 100));// Expect -1
        int[] arr2 = new int[]{10};
        System.out.println(find(arr2, 10));// Expect 0
        System.out.println(find(arr2, 100));// Expect -1

        int[] arr3 = new int[]{5, 6, 2, 4};
        System.out.println(find(arr3, 6));// Expect 1
        System.out.println(find(arr3, 9));// Expect -1
        int[] arr4 = new int[]{5, 6, 10, 100};
        System.out.println(find(arr4, 5));// Expect 0
        System.out.println(find(arr4, 101));// Expect -1
        int[] arr5 = new int[]{2, 3, 4, 5, 1};
        System.out.println(find(arr5, 1));// Expect 4
        System.out.println(find(arr5, 9));// Expect -1
    }

    public void testFindMinElement() {
        System.out.println(findMinElement(null));// Expect -1
        int[] arr1 = new int[]{};
        System.out.println(findMinElement(arr1));// Expect -1
        int[] arr2 = new int[]{10};
        System.out.println(findMinElement(arr2));// Expect 0
        int[] arr3 = new int[]{5, 6, 2, 4};
        System.out.println(findMinElement(arr3));// Expect 2
        int[] arr4 = new int[]{5, 6, 10, 100};
        System.out.println(findMinElement(arr4));// Expect 0
        int[] arr5 = new int[]{2, 3, 4, 5, 1};
        System.out.println(findMinElement(arr5));// Expect 4
    }

    public static int findMinElement(int[] arr) {
        if (arr == null || arr.length == 0) {
            return -1;
        }
        int l = 0, h = arr.length - 1;
        //If arr[l:h] sub-array is rotated, keep searching for minimum
        // if it is not rotated, arr[l] <= arr[h], you can break and return l,
        while (arr[l] > arr[h]) {
            int mid = (l + h) / 2;
            // If arr[mid] is less than its previous element arr[mid-1], We found the decreasing order
            // So we have found the start of the sorted array
            if (mid - 1 >= 0 && arr[mid] < arr[mid - 1]) {
                return mid;
            }
            //We can't find the pivot in the sorted sequence;
            //If left half is sorted, we will skip that otherwise we will skip the right half
            if (arr[l] <= arr[mid]) {
                l = mid + 1;
            } else {
                h = mid - 1;
            }
        }
        return l;
    }

    public static int find(int[] arr, int elem) {
        if (arr == null || arr.length == 0) {
            return -1;
        }
        int l = 0, h = arr.length - 1;
        while (l <= h) {
            int mid = (l + h) / 2;
            if (arr[mid] == elem)
                return mid;
            if (arr[l] <= arr[mid]) { // Left half is sorted
                if (arr[l] <= elem && arr[mid] >= elem) //Check if target element lies with in that range
                    h = mid - 1;
                else
                    l = mid + 1;
            } else { // Right half is sorted
                if (arr[mid] <= elem && arr[h] >= elem) {
                    l = mid + 1;
                } else {
                    h = mid - 1;
                }
            }
        }
        return -1; //Element not found
    }
}