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

implSolution{pubfnmax_turbulence_size(arr:Vec<i32>) -> i32{letmut count0 = 1;letmut count1 = 1;letmut ret = 1;for k in0..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)}}