L33 Search in Rotated Sorted Array

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly rotated at an unknown pivot index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be rotated at pivot index 3 and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

You must write an algorithm with O(log n) runtime complexity.

Example 1:

Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4

Example 2:

Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1

Example 3:

Input: nums = [1], target = 0
Output: -1

Constraints:

• 1 <= nums.length <= 5000
• -104 <= nums[i] <= 104
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -104 <= target <= 104

public class search_In_Rotated_Sorted_Array_L33{
    public static void main(String[] args){
        int[] nums = {4, 5, 6, 7, 0, 1, 2};
        int target = 0;
        int target2 = 3;
        int[] nums2= {1};
        int target3 = 0;
        int[] num3 = {4, 5, 6, 7, 0, 1, 2};
        System.out.println(search(nums, target));
        System.out.println(search(nums, target2));
        System.out.println(search(nums2, target3));
    }

    public static int search(int[] nums, int target){
        // base case
        if (nums == null || nums.length == 0){
            return -1;
        }
        int k = 0;
        for (int i = 1; i < nums.length; i++){
            if (nums[i] < nums[i -1]){
                k = i;
            }   
        }

        if (k == 0){
            return binarysearch(nums, target);
        }

        int[] partone = new int[k];
        for (int i = 0; i < k; i++){
            partone[i] = nums[i];
        }

        int[] parttwo = new int[nums.length - k];
        for (int i = 0; i < nums.length - k; i++){
            parttwo[i] = nums[k+i];
        }

        if (target >= nums[0]){
            return binarysearch(partone, target);
        }else {
            int result = binarysearch(parttwo, target);
            if (result != -1){
                return (result+k);
            }else{
                return result;
            }
        }
    }   

    public static int binarysearch(int[] nums, int target){
        int left = 0;
        int right = nums.length -1;
        while(left <= right){
            int mid = left + (right - left)/2;
            if (nums[mid] == target){
                   return mid;
            }else if (nums[mid] < target){
                left = mid + 1;
            }else {
                right = mid - 1;
            }
        }   
        return -1;

    }
}

Time Complexity: O(logN)

Space Complexity: O(1)