## # 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
``````

• This is a follow up problem to Search in Rotated Sorted Array, where nums may contain duplicates.
• Would this affect the run-time complexity? How and why?

### # Solution

Approach 1: binary search -- same with 0033 Search in Rotated Sorted Array, but worse case can be linear, e.g. `[0,0,0,0,0]`.

### # Code (Python)

Approach 1:

``````class Solution:
def search(self, nums, target):
"""
:type nums: List[int]
:type target: int
:rtype: int
"""
if not nums:
return False
left, right = 0, len(nums) - 1
while left < right:
mid = (left + right) // 2
if target in (nums[left], nums[mid], nums[right]):
return True
if nums[mid] > nums[right]:
# example: [4,5,6,7,0,1,2]
if nums[left] < target < nums[mid]:
right = mid - 1
else:
left = mid + 1
elif nums[mid] < nums[right]:
# example: [5,0,1,2,3]
if nums[mid] < target < nums[right]:
left = mid + 1
else:
right = mid - 1
else: # when nums[mid] == nums[right], move either pointer
right -= -1
return True if nums[left] == target else False
``````

### # Code (C++)

Approach 1:

``````class Solution {
public:
bool search(vector<int>& nums, int target) {
int tail = nums.size() - 1;
{
if (nums[mid] == target)
return true;
{
if (nums[mid] < target)
else
tail = mid - 1;
}
else
{
if (nums[mid] < target)
{
if (nums[mid] >= nums[head] || target <= nums[tail])
else if (target > nums[tail])
tail = mid - 1;
}
else if (nums[mid] > target)
{
if (nums[mid] <= nums[tail] || target >= nums[head])
tail = mid - 1;