README.md

978. Longest Turbulent Subarray

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 (Rust)

1. Solution

impl Solution {
    pub fn max_turbulence_size(arr: Vec<i32>) -> i32 {
        let mut count0 = 1;
        let mut count1 = 1;
        let mut ret = 1;

        for k in 0..arr.len() - 1 {
            if (k % 2 == 1 && arr[k] > arr[k + 1]) || (k % 2 == 0 && arr[k] < arr[k + 1]) {
                count0 += 1;
            } else {
                ret = ret.max(count0);
                count0 = 1;
            }
            if (k % 2 == 0 && arr[k] > arr[k + 1]) || (k % 2 == 1 && arr[k] < arr[k + 1]) {
                count1 += 1;
            } else {
                ret = ret.max(count1);
                count1 = 1;
            }
        }

        ret.max(count0).max(count1)
    }
}