# 33. Search in Rotated Sorted Array

## 27 Dec 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, nums, ..., 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 = , 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:
num = len(nums)
left = 0
right=len(nums) -1
if num == 0:
return -1

while left <= right:
middle = left + (right – left) //2
print(middle)
if nums[middle] == target:
return middle
# inflection point to the right. Left is strictly increasing
if nums[middle]>= nums[left]:
if nums[left] <= target < nums[middle]:
right = middle – 1
else:
left = middle +1
# inflection point to the left of me. Right is strictly increasing
else:
if nums[middle]< target <= nums[right]:
left = middle + 1
else:
right = middle -1
print(“————————“)
return -1