# 845. Longest Mountain in Array

## Description

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

• Can you solve it using only one pass?
• Can you solve it in O(1) space?

## Solutions

• class Solution {
public int longestMountain(int[] arr) {
int n = arr.length;
int ans = 0;
for (int l = 0, r = 0; l + 2 < n; l = r) {
r = l + 1;
if (arr[l] < arr[r]) {
while (r + 1 < n && arr[r] < arr[r + 1]) {
++r;
}
if (r + 1 < n && arr[r] > arr[r + 1]) {
while (r + 1 < n && arr[r] > arr[r + 1]) {
++r;
}
ans = Math.max(ans, r - l + 1);
} else {
++r;
}
}
}
return ans;
}
}

• class Solution {
public:
int longestMountain(vector<int>& arr) {
int n = arr.size();
int ans = 0;
for (int l = 0, r = 0; l + 2 < n; l = r) {
r = l + 1;
if (arr[l] < arr[r]) {
while (r + 1 < n && arr[r] < arr[r + 1]) {
++r;
}
if (r + 1 < n && arr[r] > arr[r + 1]) {
while (r + 1 < n && arr[r] > arr[r + 1]) {
++r;
}
ans = max(ans, r - l + 1);
} else {
++r;
}
}
}
return ans;
}
};

• class Solution:
def longestMountain(self, arr: List[int]) -> int:
n = len(arr)
ans = l = 0
while l + 2 < n:
r = l + 1
if arr[l] < arr[r]:
while r + 1 < n and arr[r] < arr[r + 1]:
r += 1
if r < n - 1 and arr[r] > arr[r + 1]:
while r < n - 1 and arr[r] > arr[r + 1]:
r += 1
ans = max(ans, r - l + 1)
else:
r += 1
l = r
return ans


• func longestMountain(arr []int) (ans int) {
n := len(arr)
for l, r := 0, 0; l+2 < n; l = r {
r = l + 1
if arr[l] < arr[r] {
for r+1 < n && arr[r] < arr[r+1] {
r++
}
if r+1 < n && arr[r] > arr[r+1] {
for r+1 < n && arr[r] > arr[r+1] {
r++
}
ans = max(ans, r-l+1)
} else {
r++
}
}
}
return
}