852 Peak Index in a Mountain Array

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 a mountain array arr, return the index i such that arr[0] < arr[1] < ... < arr[i - 1] < arr[i] > arr[i + 1] > ... > arr[arr.length - 1].

You must solve it in O(log(arr.length)) time complexity.

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 <= 105
• 0 <= arr[i] <= 106
• arr is guaranteed to be a mountain array.

Binary search over the array of comparisons to find the largest index i such that A[i] < A[i+1].

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