33 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`
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 `````` ``````class Solution: def search(self, nums: List[int], target: int) -> int: n = len(nums) lo, hi = 0, n-1 while lo <= hi: mid = lo + (hi - lo) // 2 if nums[mid] == target: return mid elif nums[mid] >= nums[lo]: if nums[lo] <= target < nums[mid]: hi = mid - 1 else: lo = mid + 1 else: if nums[mid] < target <= nums[hi]: lo = mid + 1 else: hi = mid - 1 return -1``````