"""

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

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?

"""


# Linear Scan

"""
Time Complexity: O(N), where N is the length of A.
Space Complexity: O(1).
"""


class Solution(object):
    def peakIndexInMountainArray(self, A):
        for i in range(len(A)):
            if A[i] > A[i+1]:
                return i

# Binary Search


"""
Time Complexity: O(logN), where N is the length of A.
Space Complexity: O(1).
"""


class Solution(object):
    def peakIndexInMountainArray(self, A):
        lo, hi = 0, len(A) - 1
        while lo < hi:
            mi = (lo + hi) // 2
            if A[mi] < A[mi + 1]:
                lo = mi + 1
            else:
                hi = mi
        return lo
