81. 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?

Since the array is given in a sorted order, so it can be solved using the binary search algorithm.

To solve this problem we have to follow the folllowing steps:

1. Calculate the mid index.
2. Check if the mid element == target, return True else move to next step.
3. Else if the mid element >= left.
   if mid element >= target and and left <= target, then shift right to mid-1 position, else shift left to mid+1 position.
4. Else,
   If target >= mid element and target <=right, then shift left to mid+1 position, else shift right to mid-1 position.
5. If the element is not found return False

Note: Since duplicate elemnts are present in the array so remove all the duplicates before step step 1.
To remove duplicate,

1. Shift left while left == left+1, and
2. Shift right while right == right-1.

# If the length of the given array list is 1, then check the first element and return accordingly
if len(nums)==1:
    if nums[0]!=target:
        return False
    else:
        return True

left=0
right=len(nums)-1
# binary search 
while(left<=right):

    # shifting to remove duplicate elements
    while left<right and nums[left] == nums[left+1]:
        left+=1
    while left<right and nums[right] == nums[right-1]:
        right-=1

    # step 1 calculate the mid    
    mid=(left+right)//2

    #step 2
    if nums[mid]==target:
        return True

    #step 3
    elif nums[left]<=nums[mid]:
        if nums[mid]>=target and nums[left]<=target:
            right=mid-1
        else:
            left=mid+1

    # step 4
    else:
        if target>=nums[mid] and target<=nums[right]:
            left=mid+1
        else:
            right=mid-1

# step 5
return False