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978. Longest Turbulent Subarray

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: 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 \(O(1)\).

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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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
    }
}

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