Problem Statement:

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
  • -10^4 <= nums[i] <= 10^4
  • All values of nums are unique.
  • nums is an ascending array that is possibly rotated.
  • -10^4 <= target <= 10^4

Solution:

class Solution {
    public int search(int[] nums, int target) {
        int start = 0;
        int end = nums.length - 1;
 
        while (start <= end) {
            int mid = start + (end - start) / 2;
 
            if (nums[mid] == target) {
                return mid;
            }
 
            // Determine which part is sorted
            if (nums[start] <= nums[mid]) {
                // Left half is sorted
                if (nums[start] <= target && target < nums[mid]) {
                    end = mid - 1;  // Target in the left half
                } else {
                    start = mid + 1;  // Target in the right half
                }
            } else {
                // Right half is sorted
                if (nums[mid] < target && target <= nums[end]) {
                    start = mid + 1;  // Target in the right half
                } else {
                    end = mid - 1;  // Target in the left half
                }
            }
        }
 
        return -1;  // Target not found
    }
}