Find guaranteed single element in duplicated sorted array

"""
You are given a sorted array consisting of only integers where every element appears exactly twice, except for one element which appears exactly once.

Return the single element that appears only once.
"""
class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        left, right = 0, len(nums)-1

        while left < right:
            mid = left + (right - left) // 2

            """
            Assuming there is no odd-one-out-element e.g [1,1,2,2,3,3],
            if mid = 0 -> its duplicate is at mid = mid + 1
            if mid = 1 -> its duplicate is at mid = mid - 1
            """
            if ( (mid % 2 == 1 and nums[mid - 1] == nums[mid]) or
                 (mid % 2 == 0 and nums[mid] == nums[mid + 1]) ):

                left = mid + 1
            else:
                right = mid

        return nums[left]