# 845 Longest Mountain in Array

You may recall that an array `arr` is a mountain array if and only if:

• `arr.length >= 3`
• There exists some index `i` (0-indexed) with `0 < i < arr.length - 1` such that:
• `arr < arr < ... < arr[i - 1] < arr[i]`
• `arr[i] > arr[i + 1] > ... > arr[arr.length - 1]`

Given an integer array `arr`, return the length of the longest subarray, which is a mountain. Return `0` if there is no mountain subarray.

Example 1:

``````Input: arr = [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: arr = [2,2,2]
Output: 0
Explanation: There is no mountain.
``````

Constraints:

• `1 <= arr.length <= 104`
• `0 <= arr[i] <= 104`

• Can you solve it in `O(1)` space?
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 `````` ``````class Solution: def longestMountain(self, arr: List[int]) -> int: n = len(arr) up, down =  * n,  * n for i in range(1, n): if arr[i] > arr[i - 1]: up[i] = up[i - 1] + 1 for i in range(n-2, -1, -1): if arr[i] > arr[i + 1]: down[i] = down[i + 1] + 1 return max( [u + d + 1 for u, d in zip(up, down) if u and d] or  ) ''' One pass ''' class Solution: def longestMountain(self, arr: List[int]) -> int: n = len(arr) res = up = down = 0 for i in range(1, n): if ( down and arr[i - 1] < arr[i] or arr[i - 1] == arr[i] ): up = down = 0 up += arr[i - 1] < arr[i] down += arr[i - 1] > arr[i] if up and down: res = max(res, up + down + 1) return res``````