LeetCode-in-Java

978. Longest Turbulent Subarray

Medium

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

Solution

public class Solution {
    public int maxTurbulenceSize(int[] arr) {
        int n = arr.length;
        int ans = 1;
        int l;
        int r;
        if (n == 1) {
            return 1;
        }
        if (n == 2) {
            return arr[0] == arr[1] ? 1 : 2;
        }
        for (l = 0, r = 1; r < n - 1; r++) {
            int difL = arr[r] - arr[r - 1];
            int difR = arr[r] - arr[r + 1];
            if (difL == 0 && difR == 0) {
                l = r + 1;
            } else if (difL == 0) {
                ans = Math.max(ans, r - l);
                l = r;
            } else if (!((difL < 0 && difR < 0) || (difL > 0 && difR > 0))) {
                ans = Math.max(ans, r - l + 1);
                l = r;
            }
        }
        return Math.max(ans, r - l + 1);
    }
}

This site is open source. Improve this page.