# Search in Rotated Sorted Array

Updated: Mar 25, 2021

There is an integer array nums sorted in ascending order (with **distinct** values).

Prior to being passed to your function, nums is **rotated** at an unknown pivot index k (0 <= 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 rotation and an integer target, return *the index of *target* if it is in *nums*, or *-1* if it is not in *nums.

**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 guaranteed to be rotated at some pivot.

-104 <= target <= 104

** Solution:**

```
class Solution {
public int search(int[] nums, int target) {
int left =0,right=nums.length-1;
int mid;
while(left<=right)
{
mid=(left+right)/2;
if(nums[mid]==target) return mid;
else if(nums[mid]>=nums[left])
{
if(target<=nums[mid] && target>=nums[left])
right = mid-1;
else
left=mid+1;
}
else
{
if(target>=nums[mid] && target<=nums[right])
left=mid+1;
else
right=mid-1;
}
}
return -1;
}
}
```