5. 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

Approach

• Compute mid then check if it matches target then return mid
• Check if mid is in left sorted or right sorted
• For left check if it is out -> like target > mid or target < l then go left else right
• For right check if it is out -> target < mid or target > rigth

class Solution {
    public int search(int[] nums, int target) {
        int l = 0, r = nums.length - 1;
 
        while (l <= r) {
            int m = l + (r-l) / 2;
            if (target == nums[m]) return m;
 
            if (nums[l] <= nums[m]) {
                if (target > nums[m] || target < nums[l])
                    l = m + 1;
                else
                    r = m - 1;    
            } else {
                if (target < nums[m] || target > nums[r])
                    r = m - 1;
                else
                    l = m + 1;    
            }
        }
        return -1;
    }
}