Peak Index in a Mountain Array

852. Peak Index in a Mountain Array | Easy

The solution contained is naive approach and can be improved much better.

Problem:

Let's call an array A a mountain if the following properties hold:

A.length >= 3
There exists some 0 < i < A.length - 1 such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1]
Given an array that is definitely a mountain, return any i such that A[0] < A[1] < ... A[i-1] < A[i] > A[i+1] > ... > A[A.length - 1].

Example 1:

Input: [0,1,0]
Output: 1
Example 2:

Input: [0,2,1,0]
Output: 1
Note:

3 <= A.length <= 10000
0 <= A[i] <= 10^6
A is a mountain, as defined above.

Accepted 47,112 Submissions 69,111

My attempt

class Solution:
    def peakIndexInMountainArray(self, A):
        """
        :type A: List[int]
        :rtype: int
        """

        """
        # initial approach - naive solution
        # 108ms, 0.88% on speed
        _index = 0
        _elem = 0
        for index, elem in enumerate(A):
            if elem > _elem:
                _index = index
                _elem = elem
        return _index
        """