33. Search in Rotated Sorted Array

Medium

---

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

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        pivot = self.findPivot(nums, target)
        return pivot
    
    def findPivot(self, nums, target):
        low = 0
        high = len(nums)-1
        while low <= high:
            mid = low+(high-low)//2
            if nums[mid] == target:
                return mid
            elif nums[low] <= nums[mid]:
                if nums[low] <= target <= nums[mid]:
                    high = mid - 1
                else:
                    low = mid + 1
            else:
                if nums[mid] <= target <= nums[high]:
                    low = mid + 1
                else:
                    high = mid - 1
        return -1