# 81. 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?

```cpp
// Binary Search
bool search(vector<int>& nums, int target) { // time: O(logn); space: O(1)
    int left = 0, right = nums.size() - 1;
    while (left <= right) {
        int mid = left + (right - left) / 2;
        if (nums[mid] == target) return true;
        if (nums[left] < nums[mid]) { // 1st half is in order
            if (nums[left] <= target && target < nums[mid]) {
                right = mid - 1;
            } else {
                left = mid + 1;
            }
        } else if (nums[left] > nums[mid]) { // 2nd half is in order
            if (nums[mid] < target && target <= nums[right]) {
                left = mid + 1;
            } else {
                right = mid - 1;
            }
        } else { // handle the case like: [3, 1, 2, 3, 3, 3]
            ++left;
        }
    }
    return false;
}
```

```python
class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        left = 0
        right = len(nums) - 1
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] == target:
                return True
            if nums[left] < nums[mid]:
                if nums[left] <= target and target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            elif nums[left] > nums[mid]:
                if nums[mid] < target and target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
            else:
                left = left + 1
        return False
```

{% content-ref url="33.-search-in-rotated-sorted-array" %}
[33.-search-in-rotated-sorted-array](https://jimmylin1991.gitbook.io/practice-of-algorithm-problems/binary-search/33.-search-in-rotated-sorted-array)
{% endcontent-ref %}