# 81. Search in Rotated Sorted Array II

#### Medium

***

There is an integer array `nums` sorted in non-decreasing order (not necessarily with **distinct** values).

Before 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, `[0,1,2,4,4,4,5,6,6,7]` might be rotated at pivot index `5` and become `[4,5,6,6,7,0,1,2,4,4]`.

Given the array `nums` **after** the rotation and an integer `target`, return `true` *if* `target` *is in* `nums`*, or* `false` *if it is not in* `nums`*.*

You must decrease the overall operation steps as much as possible.

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**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
```

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**Constraints:**

* `1 <= nums.length <= 5000`
* `-104 <= nums[i] <= 104`
* `nums` is guaranteed to be rotated at some pivot.
* `-104 <= target <= 104`

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**Follow up:** This problem is similar to Search in Rotated Sorted Array, but `nums` may contain **duplicates**. Would this affect the runtime complexity? How and why?

**Answer** : In Worst case all the elements can be equal in that case binary search will not help and in that case we would be simply scanning all elements one by one (**Point 1 in code**). So Complexity in worst case will be ***O(N)***

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