Single Element in a Sorted Array

'''
Pairs start at even index: (0,1), (2,3), (4,5)... so at any even place number for eg. i will always have one similary number at i+1 if all the elements are having 2 occurance from 0 to i+1.

Steps:
find mid and check if it is even or odd, if it odd, make it even first by reducing 1 from it.
if arr[mid]==arr[mid+1] then left single occurance number is in right because left have all elements in pair so after mid+1 you
if arr[mid]!=arr[mid+1] means arr[mid] is the element of arr[mid]==arr[mid-1] means element is in left so check in the left part of the array.

steps:
1.find mid and if mid if odd,make it even.
2.if mid and mid+1th element is same, adjust l to mid+2 else r=mid

Tricks:
1.Keep l<r because we need that midth element.
2.when you find the single frequent element is in right side so make l=mid+2 else it can on infinite loop for array=[1,1,2] l=0, r=2, mid= 1 and mid will get adjusted to 0 to make it even so it will stay forever at 0 so to find the element,make it right.
'''
class Solution:
    def singleNonDuplicate(self, nums: List[int]) -> int:
        l,r=0,len(nums)-1
        while l<r: #l=0, r=8, mid=4
            mid = (l+r)//2
            if mid%2==1:
                mid-=1
            if nums[mid]==nums[mid+1]:
                l=mid+2 #mid and mid+1 element is same so move to the left boundary to the next pair.
            else:
                r=mid
        return nums[l]