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

 

Constraints:

  • 3 <= arr.length <= 104
  • 0 <= arr[i] <= 106
  • arr is guaranteed to be a mountain array.

 

Follow up: Finding the O(n) is straightforward, could you find an O(log(n)) solution?SOLUTION:
class Solution:
    def peakIndexInMountainArray(self, arr: List[int]) -> int:
        old = True
        new = True
        for i in range(len(arr) - 1):
            new = arr[i + 1] > arr[i]
            if new ^ old:
                return i
            old = new

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