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: 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 Follow up: Can you solve it using only one pass? Can you solve it in O(1) space?