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 {
public:
    int findPeakElement(vector<int>& nums) {
        if(nums.size()==1) return 0;
        int i=0,j=nums.size()-1;
        int n=nums.size();
        while(i<j){
            int mid=(i+j)/2;
            if((mid==0 || nums[mid-1]<nums[mid]) && (mid==n || nums[mid+1]<nums[mid])) return mid;
            if(nums[mid+1]>nums[mid]) i=mid+1;
            else j=mid-1;
        }
        return i;
    }
};

#Python
class Solution:
    def findPeakElement(self, nums: List[int]) -> int:
        if len(nums)==1:
            return 0
        l,r=0,len(nums)-1
        while l<r:
            mid=(l+r)//2
            print(mid)
            if (mid==0 or nums[mid-1]<nums[mid]) and (mid==len(nums) or nums[mid]>nums[mid+1]):
                return mid
            elif mid>0 and nums[mid-1]>nums[mid]:
                r=mid-1
            elif mid<len(nums):
                l=mid+1
        return l