# 978. Longest Turbulent Subarray

## Description

Given an integer array arr, return the length of a maximum size turbulent subarray of arr.

A subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray.

More formally, a subarray [arr[i], arr[i + 1], ..., arr[j]] of arr is said to be turbulent if and only if:

• For i <= k < j:
• arr[k] > arr[k + 1] when k is odd, and
• arr[k] < arr[k + 1] when k is even.
• Or, for i <= k < j:
• arr[k] > arr[k + 1] when k is even, and
• arr[k] < arr[k + 1] when k is odd.

Example 1:

Input: arr = [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: arr[1] > arr[2] < arr[3] > arr[4] < arr[5]


Example 2:

Input: arr = [4,8,12,16]
Output: 2


Example 3:

Input: arr = [100]
Output: 1


Constraints:

• 1 <= arr.length <= 4 * 104
• 0 <= arr[i] <= 109

## Solutions

• class Solution {
public int maxTurbulenceSize(int[] arr) {
int ans = 1, f = 1, g = 1;
for (int i = 1; i < arr.length; ++i) {
int ff = arr[i - 1] < arr[i] ? g + 1 : 1;
int gg = arr[i - 1] > arr[i] ? f + 1 : 1;
f = ff;
g = gg;
ans = Math.max(ans, Math.max(f, g));
}
return ans;
}
}

• class Solution {
public:
int maxTurbulenceSize(vector<int>& arr) {
int ans = 1, f = 1, g = 1;
for (int i = 1; i < arr.size(); ++i) {
int ff = arr[i - 1] < arr[i] ? g + 1 : 1;
int gg = arr[i - 1] > arr[i] ? f + 1 : 1;
f = ff;
g = gg;
ans = max({ans, f, g});
}
return ans;
}
};

• class Solution:
def maxTurbulenceSize(self, arr: List[int]) -> int:
ans = f = g = 1
for a, b in pairwise(arr):
ff = g + 1 if a < b else 1
gg = f + 1 if a > b else 1
f, g = ff, gg
ans = max(ans, f, g)
return ans


• func maxTurbulenceSize(arr []int) int {
ans, f, g := 1, 1, 1
for i, x := range arr[1:] {
ff, gg := 1, 1
if arr[i] < x {
ff = g + 1
}
if arr[i] > x {
gg = f + 1
}
f, g = ff, gg
ans = max(ans, max(f, g))
}
return ans
}

• function maxTurbulenceSize(arr: number[]): number {
let f = 1;
let g = 1;
let ans = 1;
for (let i = 1; i < arr.length; ++i) {
const ff = arr[i - 1] < arr[i] ? g + 1 : 1;
const gg = arr[i - 1] > arr[i] ? f + 1 : 1;
f = ff;
g = gg;
ans = Math.max(ans, f, g);
}
return ans;
}