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# 000033-Search-in-Rotated-Sorted-Array

### Problem

<https://leetcode.com/problems/search-in-rotated-sorted-array/description/>

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

### Solution

binary search

```python
class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if not nums:
            return -1
        
        left, right = 0, len(nums) - 1
        while left < right:
            mid = (left+right)//2
            if nums[mid] == target:
                return mid
            if nums[mid] >= nums[left]:
                if nums[left] <= target <= nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                if nums[mid] <= target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
        return left if nums[left] == target else -1
```