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# Search in Rotated Sorted Array II

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., `[0,0,1,2,2,5,6]` might become `[2,5,6,0,0,1,2]`).

You are given a target value to search. If found in the array return `true`, otherwise return `false`.

**Example 1:**

```
Input: nums = [2,5,6,0,0,1,2], target = 0
Output: true
```

**Example 2:**

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

**Follow up:**

* This is a follow up problem to [Search in Rotated Sorted Array](https://leetcode.com/problems/search-in-rotated-sorted-array/description/), where `nums` may contain duplicates.
* Would this affect the run-time complexity? How and why?

```java
class Solution {
    public boolean search(int[] nums, int target) {
        int start = 0, end = nums.length - 1, mid = -1;
        while (start <= end) {
            mid = start + (end - start) / 2;
            if (nums[mid] == target)
                return true;
            // If we know for sure right side is sorted or left side is unsorted
            if (nums[mid] < nums[end] || nums[mid] < nums[start])
                if (target > nums[mid] && target <= nums[end])
                    start = mid + 1;
                else
                    end = mid - 1;
            // If we know for sure left side is sorted or right side is unsorted
            else if (nums[mid] > nums[start] || nums[mid] > nums[end])
                if (target < nums[mid] && target >= nums[start])
                    end = mid - 1;
                else
                    start = mid + 1;
            // If we get here, that means nums[start] == nums[mid] == nums[end], then shifting out
            // any of the two sides won't change the result but can help remove duplicate from
            // consideration, here we just use end-- but left++ works too
            else
                end--;
        }
        return false;
    }
}
```