README_EN.md

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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:

  <ul>
  	<li><code>arr[k] &gt; arr[k + 1]</code> when <code>k</code> is odd, and</li>
  	<li><code>arr[k] &lt; arr[k + 1]</code> when <code>k</code> is even.</li>
  </ul>
  </li>
  <li>Or, for <code>i &lt;= k &lt; j</code>:
  <ul>
  	<li><code>arr[k] &gt; arr[k + 1]</code> when <code>k</code> is even, and</li>
  	<li><code>arr[k] &lt; arr[k + 1]</code> when <code>k</code> is odd.</li>
  </ul>
  </li>
  

 

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: Dynamic Programming

We define $f[i]$ as the length of the longest turbulent subarray ending at $\textit{nums}[i]$ with an increasing state, and $g[i]$ as the length of the longest turbulent subarray ending at $\textit{nums}[i]$ with a decreasing state. Initially, $f[0] = 1$, $g[0] = 1$. The answer is $\max(f[i], g[i])$.

For $i \gt 0$, if $\textit{nums}[i] \gt \textit{nums}[i - 1]$, then $f[i] = g[i - 1] + 1$, otherwise $f[i] = 1$; if $\textit{nums}[i] \lt \textit{nums}[i - 1]$, then $g[i] = f[i - 1] + 1$, otherwise $g[i] = 1$.

Since $f[i]$ and $g[i]$ are only related to $f[i - 1]$ and $g[i - 1]$, two variables can be used instead of arrays.

The time complexity is $O(n)$, where $n$ is the length of the array. The space complexity is $O(1)$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}

Rust

impl Solution {
    pub fn max_turbulence_size(arr: Vec<i32>) -> i32 {
        let mut ans = 1;
        let mut f = 1;
        let mut g = 1;

        for i in 1..arr.len() {
            let ff = if arr[i - 1] < arr[i] { g + 1 } else { 1 };
            let gg = if arr[i - 1] > arr[i] { f + 1 } else { 1 };
            f = ff;
            g = gg;
            ans = ans.max(f.max(g));
        }

        ans
    }
}