# Search in Rotated Sorted Array Leetcode Solution

In this post, we are going to solve the Search in Rotated Sorted Array Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

## Problem

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`

Now, lets see the leetcode solution of Search in Rotated Sorted Array Leetcode Solution.

### Search in Rotated Sorted Array Leetcode Solution in Python

```class Solution:

def search(self, nums, target):
if not nums:
return -1

low, high = 0, len(nums) - 1

while low <= high:
mid = (low + high) / 2
if target == nums[mid]:
return mid

if nums[low] <= nums[mid]:
if nums[low] <= target <= nums[mid]:
high = mid - 1
else:
low = mid + 1
else:
if nums[mid] <= target <= nums[high]:
low = mid + 1
else:
high = mid - 1

return -1```

### Search in Rotated Sorted ArrayLeetcode Solutionin CPP

```class Solution {
public:
int search(vector<int>& nums, int target) {
int s = 0;
int e = nums.size()-1;
while(s<=e)
{
int mid = (s+e)>>1;
if(target==nums[mid])return mid;
else if(nums[mid]<=nums[e])
{if(target>=nums[mid] and target<=nums[e])
s=mid+1;
else e = mid-1;
}
else{
if(target>=nums[s] and target<=nums[mid])
e = mid-1;
else s = mid+1;
}
}
return -1;
}
};```

### Search in Rotated Sorted Array Leetcode Solution in Java

```class Solution {
public int search(int[] nums, int target) {

int s = 0;
int e = nums.length-1;

while(s <= e){
int mid = s - (s - e)/2;
if(nums[mid] == target) return mid;

if(nums[s] <= nums[mid]){
if(target >= nums[s] && target <= nums[mid])  e = mid - 1;
else s = mid+1;
}
if(nums[mid] <= nums[e]){
if(target <= nums[e] && target >= nums[mid])  s = mid + 1;
else e = mid - 1;
}
}

return -1;
}
}```

Note: This problem Search in Rotated Sorted Array is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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