LeetCode 0978. Longest Turbulent Subarray Solution in Java, C++, Python & More | Explanation + Code

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 nums[i] with an increasing state, and g[i] as the length of the longest turbulent subarray ending at 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 nums[i] \gt nums[i - 1], then f[i] = g[i - 1] + 1, otherwise f[i] = 1; if nums[i] \lt 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).

PythonJavaC++GoTypeScriptRust

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(code-box)

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; } }(code-box)

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; } };(code-box)

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 }(code-box)

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; }(code-box)

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 } }(code-box)