# Search in Rotated Sorted Array – Leetcode Python

LeetCode has a Medium coding Problem in Its’ Algorithm Section “Search in Rotated Sorted Array”. Today We are going to solve this problem. LeetCode Link of the Problem is HERE

#### Question

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.

Examples

```Input: nums = [4,5,6,7,0,1,2], target = 0
Output: 4
```
```Input: nums = [4,5,6,7,0,1,2], target = 3
Output: -1
```
```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`

#### Solution to Search in Rotated Sorted Array

Skeleton code

``````class Solution(object):
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""``````

If you find the target value in the array return its index, otherwise return -1. You may assume no duplicate exists in the array.

Complete Solution

``````class Solution:
def search(self, nums, target):
l, r = 0, len(nums)-1
while l <= r:
mid = l + (r-l)//2
if nums[mid] == target:
return mid
if nums[l] <= nums[mid]:  # here should include "==" case
if nums[l] <= target < nums[mid]:
r = mid - 1
else:
l = mid + 1
else:
if nums[mid] < target <= nums[r]:
l = mid + 1
else:
r = mid - 1
return -1``````

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