# LeetCode 33. Search in Rotated Sorted Array

## Description

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

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., `[0,1,2,4,5,6,7]` might become `[4,5,6,7,0,1,2]`).

You are given a target value to search. If found in the array return its index, otherwise return `-1`.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

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```

## Explanation

find the rotation point and then do binary search

## Python Solution

``````class Solution:
def search(self, nums: List[int], target: int) -> int:
result = -1
if nums == None or len(nums) == 0:
return -1

min_index = self.find_min_index(nums)

if nums[min_index] <= target <= nums[-1]:
return self.binary_search(nums, target, min_index, len(nums) - 1)

return self.binary_search(nums, target, 0, min_index)

def find_min_index(self, nums):
start = 0
end = len(nums) - 1

while start + 1 < end:
mid = start + (end - start) // 2

if nums[mid] < nums[end]:
end = mid
else:
start = mid

if nums[start] < nums[end]:
return start

return end

def binary_search(self, nums, target, start, end):
while start + 1 < end:
mid = start + (end - start) // 2

if nums[mid] == target:
return mid
elif nums[mid] < target:
start = mid
else:
end = mid

if nums[start] == target:
return start
elif nums[end] == target:
return end

return -1``````
• Time complexity: O(N).
• Space complexity: O(1).