## DEV Community

Abhishek Chaudhary

Posted on

# 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`

SOLUTION:

``````class Solution:
def getPivot(self, nums):
n = len(nums)
beg = 0
end = n
while beg <= end:
mid = (beg + end) // 2
if mid < n - 1 and nums[mid] > nums[mid + 1]:
return mid + 1
if beg >= end - 1:
return 0
elif nums[mid] > nums[0]:
beg = mid
else:
end = mid

def binarySearch(self, nums, beg, end, target):
while beg <= end:
mid = (beg + end) // 2
if mid >= end:
break
elif nums[mid] == target:
return mid
elif beg == end:
break
elif nums[mid] > target:
end = mid
else:
beg = mid + 1
return -1

def search(self, nums: List[int], target: int) -> int:
n = len(nums)
k = self.getPivot(nums)
left = self.binarySearch(nums, 0, k, target)
if left != -1:
return left
right = self.binarySearch(nums, k, n, target)
if right != -1:
return right
return -1
``````