1020. Longest Turbulent Subarray¶

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1020. Longest Turbulent Subarray

Medium

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

Solution¶

class Solution {
    public int maxTurbulenceSize(int[] arr) {
        int n = arr.length;
        if (n == 1) return 1;
        int count = 1, maxi = 0;
        for (int i = 0; i < n - 1; i++) {
            if (i % 2 == 0 && arr[i] > arr[i + 1]) count++;
            else if (i % 2 == 1 && arr[i] < arr[i + 1]) count++;
            else {
                maxi = Math.max(maxi, count);
                count = 1;
            }
        }
        maxi = Math.max(maxi, count);
        count = 1;
        for (int i = 0; i < n - 1; i++) {
            if (i % 2 == 0 && arr[i] < arr[i + 1]) count++;
            else if (i % 2 == 1 && arr[i] > arr[i + 1]) count++;
            else {
                maxi = Math.max(maxi, count);
                count = 1;
            }
        }
        maxi = Math.max(maxi, count);
        return maxi;
    }
}

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here