# Time:  O(logn)
# Space: O(1)

# 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.


class Solution(object):
    def peakIndexInMountainArray(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        left, right = 0, len(A)
        while left < right:
            mid = left + (right-left)//2
            if A[mid] > A[mid+1]:
                right = mid
            else:
                left = mid+1
        return left

Create peak-index-in-a-mountain-array.py