Given an array A of integers, return true if and only if it is a valid mountain array.
Recall that A is a mountain array if and only if:
A.length >= 3
There exists some i with 0 < i < A.length - 1 such that:
A[0] < A[1] < ... A[i-1] < A[i] 
A[i] > A[i+1] > ... > A[A.length - 1]
Example 1:
Input: [2,1]
Output: false
Example 2:
Input: [3,5,5]
Output: false
Example 3:
Input: [0,3,2,1]
Output: true
Note:
0 <= A.length <= 10000
0 <= A[i] <= 10000
class Solution:
    def validMountainArray(self, A: List[int]) -> bool:
        arr = A
        if len(arr) < 3:
            return False
        if arr[0] > arr[1]:
            return False
        if arr[-1] > arr[-2]:
            return False
        
        peek = None
        for i in range(1, len(arr)-1):
            if arr[i] > arr[i+1]:
                peek = i
            if peek and arr[i] < arr[i+1]:
                return False
            if arr[i] == arr[i+1] or arr[i] == arr[i-1]:
                return False
        return True
class Solution:
    def validMountainArray(self, A: List[int]) -> bool:
        arr = A
        if len(arr) < 3:
            return False
        if arr[0] > arr[1]:
            return False
        if arr[-1] > arr[-2]:
            return False
        
        peek = None
        for i in range(1, len(arr)-1):
            if arr[i] > arr[i+1]:
                peek = i
            if peek and arr[i] < arr[i+1]:
                return False
            if arr[i] == arr[i+1] or arr[i] == arr[i-1]:
                return False
        return True