## DEV Community

Abhishek Chaudhary

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# 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
``````