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.
Input: arr = [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: arr[1] > arr[2] < arr[3] > arr[4] < arr[5]
Input: arr = [4,8,12,16]
Output: 2
Input: arr = [100]
Output: 1
Reference: https://leetcode.com/explore/challenge/card/september-leetcoding-challenge-2021/638/week-3-september-15th-september-21st/3976/
int maxTurbulenceSize(vector<int>& arr) {
int result = 1, prevEle = 0;
int j = 0;
for (int i = 1; i < arr.size(); ++i) {
int differ = arr[i] - arr[i-1];
if (differ == 0) j = i;
else if ((long)prevEle * differ > 0) j = i-1;
result = max(result, i - j + 1);
prevEle = differ;
}
return result;
}