# Search in Rotated Sorted Array

## Description

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&#x20;
>
> **Output:** 4

**Example 2:**

> **Input**: nums = \[4,5,6,7,0,1,2], target = 3&#x20;
>
> **Output:** -1

## **Code**

```java
class Solution {
    public int search(int[] nums, int target) {
        if (nums.length == 0) return -1;
        int minIdx = findMinIdx(nums);
        if (target == nums[minIdx]) return minIdx;
        int m = nums.length;
        int start = (target <= nums[m - 1]) ? minIdx : 0;
        int end = (target > nums[m - 1]) ? minIdx : m - 1;

        while (start <= end) {
            int mid = start + (end - start) / 2;
            if (nums[mid] == target) return mid;
            else if (target > nums[mid]) start = mid + 1;
            else end = mid - 1;
        }
        return -1;
    }

    public int findMinIdx(int[] nums) {
        int start = 0, end = nums.length - 1;
        while (start < end) {
            int mid = start + (end -  start) / 2;
            if (nums[mid] > nums[end]) start = mid + 1;
            else end = mid;
        }
        return start;
    }
}
```
