Let's call any (contiguous) subarray B (of A) a mountain if the following properties hold:
B.length >= 3
There exists some 0 < i < B.length - 1 such that B[0] < B[1] < ... B[i-1] < B[i] > B[i+1] > ... > B[B.length - 1]
(Note that B could be any subarray of A, including the entire array A.)
Given an array A of integers, return the length of the longest mountain. 
Return 0 if there is no mountain.
Example 1:
Input: [2,1,4,7,3,2,5]
Output: 5
Explanation: The largest mountain is [1,4,7,3,2] which has length 5.
Example 2:
Input: [2,2,2]
Output: 0
Explanation: There is no mountain.
Note:
0 <= A.length <= 10000
0 <= A[i] <= 10000
Follow up:
Can you solve it using only one pass?
Can you solve it in O(1) space?
class Solution:
    def longestMountain(self, a: List[int]) -> int:
        ml = 0
        i = 0
        while(i<len(a)):
            c = 0
            if i+1<len(a) and a[i+1]>a[i]:
                c = 1
                while(i+1<len(a) and a[i+1]>a[i]):
                    c = c + 1
                    i = i + 1
                while(i+1<len(a) and a[i+1]<a[i]):
                    c = c + 1
                    ml = max(c,ml)
                    i = i + 1
            else:
                i = i + 1
        return ml
                    

class Solution:
    def longestMountain(self, a: List[int]) -> int:
        ml = 0
        i = 0
        while(i<len(a)):
            c = 0
            if i+1<len(a) and a[i+1]>a[i]:
                c = 1
                while(i+1<len(a) and a[i+1]>a[i]):
                    c = c + 1
                    i = i + 1
                while(i+1<len(a) and a[i+1]<a[i]):
                    c = c + 1
                    ml = max(c,ml)
                    i = i + 1
            else:
                i = i + 1
        return ml
                    
