Question

978. Longest Turbulent Subarray

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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

Solution 1

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

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;
  }
};

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;
}