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[0] < arr[1] < ... < 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:5Explanation:The largest mountain is [1,4,7,3,2] which has length 5.

**Example 2:**

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

**Constraints:**

`1 <= arr.length <= 10`

^{4}`0 <= arr[i] <= 10`

^{4}

**Follow up:**

- Can you solve it using only one pass?
- Can you solve it in
`O(1)`

space?

class Solution {
public int longestMountain(int[] arr) {
}
}