human
I found an interesting problem on https://leetcode.com/problems/peak-index-in-a-mountain-array/:
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.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is the Python code you requested.

class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        low = 0
        high = len(arr) - 1
        
        while low < high:
            mid1 = (low + high)//2
            mid2 = mid1 + 1
            
            if arr[mid1] <= arr[mid2]:
                low = mid2
            else:
                high = mid1
        return low