Let's call an array arr a mountain if the following properties hold:
arr.length >= 3
There exists some i with 0 < i < arr.length - 1 such that:
arr[0] < arr[1] < ... arr[i-1] < arr[i] 
arr[i] > arr[i+1] > ... > arr[arr.length - 1]
Given an integer array arr that is guaranteed to be a mountain, return any i such that arr[0] < arr[1] < ... arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].
Example 1:
Input: arr = [0,1,0]
Output: 1
Example 2:
Input: arr = [0,2,1,0]
Output: 1
Example 3:
Input: arr = [0,10,5,2]
Output: 1
Example 4:
Input: arr = [3,4,5,1]
Output: 2
Example 5:
Input: arr = [24,69,100,99,79,78,67,36,26,19]
Output: 2
Constraints:
3 <= arr.length <= 104
0 <= arr[i] <= 106
arr is guaranteed to be a mountain array.
class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        l,r=0,len(arr)-1
        while(l<=r):
            mid = l + (r - l) // 2; 
            print(mid)
            if(arr[mid-1]<arr[mid]>arr[mid+1]):
                return mid
            elif(arr[mid-1]>arr[mid]):
                r=r-1
            else:
                l=l+1

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        l,r=0,len(arr)-1
        while(l<=r):
            mid = l + (r - l) // 2; 
            print(mid)
            if(arr[mid-1]<arr[mid]>arr[mid+1]):
                return mid
            elif(arr[mid-1]>arr[mid]):
                r=r-1
            else:
                l=l+1
