Find target in Rotated Sorted Array Binary Search

#If there are no duplicates.There is second part which have duplicates. Go down and check
class Solution:
    #Always the inflection point exists in unsorted part of array.
    #Either left half or right half of inflecton point will be in increasing order surely.
    def search(self, arr: List[int], target: int) -> int:
        n=len(arr)
        l,r=0,n-1
        while l<=r:
            m=(l+r)//2
            if arr[m]==target:#target condition
                return m
            elif arr[l]<=arr[m]:#Left is sorted
                if arr[l]<=target<=arr[m]: #check if available in sorted part
                    r=m-1
                else:#If not available in sorted part move l to right side
                    l=m+1
            else:#Right is sorted
                if arr[m]<=target<=arr[r]:#check if available in sorted par
                    l=m+1
                else:#If not available in sorted part move r to left side
                    r=m-1

        return -1
#If there are  duplicates
class Solution:
    def search(self, nums: List[int], target: int) -> bool:
        n=len(nums)
        l,r=0,n-1
        while l<=r:
            m=(l+r)//2
            if nums[m]==target:
                return True
            elif nums[l]==nums[m]==nums[r]:#extra condition or else we cant judge where to go
                l+=1
                r-=1
            elif nums[l]<=nums[m]:
                if nums[l]<=target<nums[m]:
                    r=m-1
                else:
                    l=m+1
            else:
                if nums[m]<target<=nums[r]:
                    l=m+1
                else:
                    r=m-1
        return False
e.g.[1,0,1,1,1]
search=0
If we dont use that extra condition we cant judge which side to go sometimes when a[l] and a[m] are equal