## DEV Community

Abhishek Chaudhary

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# Search in Rotated Sorted Array II

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.

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

Constraints:

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

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?

SOLUTION:

``````class Solution:
def search(self, nums: List[int], target: int) -> bool:
# Initilize two pointers
begin = 0
end = len(nums) - 1
while begin <= end:
mid = (begin + end)//2
if nums[mid] == target:
return True
if nums[mid] == nums[end]: # Fail to estimate which side is sorted
end -= 1  # In worst case: O(n)
elif nums[mid] > nums[end]: # Left side of mid is sorted
if  nums[begin] <= target and target < nums[mid]: # Target in the left side
end = mid - 1
else: # in right side
begin = mid + 1
else: # Right side is sorted
if  nums[mid] < target and target <= nums[end]: # Target in the right side
begin = mid + 1
else: # in left side
end = mid - 1
return False
``````