Find peak element

'''
One observation:arr[i] can't be same as arr[i-1] && arr[i+1] for all possible i.see constraints.
Why binary search working?
One thing for sure,If an array doesn't contain a peek then it must be strictly increasing or decreasing
that is not possible here because nums[-1] and nums[n] is INT_MIN so there is definitely a peek there.

case 1: If arr[mid-1]<arr[mid] && arr[mid+1]<arr[mid] then return mid;
case 2: If arr[mid-1]>arr[mid] then arr[-1] is also smaller than arr[mid-1] then there must be some peek
        in arr[-1] to arr[m1]
case 3: vice versa
so go in the direction of side which is having greater element than arr[mid]
'''

class Solution:
    def findPeakElement(self, arr: List[int]) -> int:
        l,r=0,len(arr)-1
        while l<=r:
            mid=(l+r)//2
            print(l,r,mid)
            if (mid-1<0 or arr[mid]>arr[mid-1]) and (mid+1>=len(arr) or arr[mid]>arr[mid+1]):
                return mid
            elif (mid-1>=0 and arr[mid-1]>arr[mid]): ###no Corner case handling here, must move in the larger neighordirection
                r=mid-1
            else:
                l=mid+1
        return l