# LeetCode 852. Peak Index in a Mountain Array

## Description

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 < arr < ... 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 < arr < ... 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.

Follow up: Finding the `O(n)` is straightforward, could you find an `O(log(n))` solution?

## Explanation

We can use binary search to find peak element where it is greater than its neighbor elements on both left and right side.

## Python Solution

``````class Solution:
def peakIndexInMountainArray(self, A: List[int]) -> int:
if not A:
return -1

start = 0
end = len(A) - 1

while start + 1 < end:
mid = start + (end - start) // 2

if A[mid] > A[mid - 1] and A[mid] > A[mid + 1]:
return mid
elif A[mid] < A[mid + 1]:
start = mid
else:
end = mid

if A[end] > A[start]:
return end
else:
return start

return -1``````
• Time complexity: ~log(N)
• Space complexity: ~1