Find the Longest Turbulent Subarray
Last Updated : 07 Jun, 2024
Given an integer array arr[] of size n. A subarray is turbulent if the comparison sign flips between each adjacent pair of elements in the subarray. The task is to find the length of a maximum size turbulent subarray of arr[]. 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:
Input: arr = [9,4,2,10,7,8,8,1,9]
Output: 5
Explanation: arr[1] > arr[2] < arr[3] > arr[4] < arr[5]
Input: arr = [4,8,12,16]
Output: 2
Approach:
To solve this problem, we can use a sliding window approach. We maintain two pointers that define the current subarray. As we iterate through the array, we check if the current element continues the turbulent property with the previous element. If it does, we extend the window. If it does not, we reset the window. We keep track of the maximum length of such subarrays during this process.
Steps-by-step approach:
- Initialize two pointers, start and end, both pointing to the beginning of the array.
- Iterate through the array using the end pointer.
- Check if the current subarray maintains the turbulent condition.
- If it does, update the maximum length.
- If it does not, move the start pointer to the next position and continue.
- Return the maximum length of turbulent subarrays found.
Below is the implementation of the above approach:
C++ #include <bits/stdc++.h> using namespace std; int maxTurbulenceSize(vector<int>& arr) { int n = arr.size(); if (n < 2) return n; int maxLen = 1; int start = 0; for (int end = 1; end < n; ++end) { int cmp = arr[end - 1] == arr[end] ? 0 : (arr[end - 1] < arr[end] ? 1 : -1); if (end == n - 1 || cmp * (arr[end] == arr[end + 1] ? 0 : (arr[end] < arr[end + 1] ? 1 : -1)) != -1) { if (cmp != 0) { maxLen = max(maxLen, end - start + 1); } start = end; } } return maxLen; } int main() { // Example 1 vector<int> arr1 = {9, 4, 2, 10, 7, 8, 8, 1, 9}; cout << maxTurbulenceSize(arr1) << endl; // Output: 5 return 0; } //code by flutterfly public class TurbulenceSize { public static int maxTurbulenceSize(int[] arr) { int n = arr.length; if (n < 2) return n; int maxLen = 1; int start = 0; for (int end = 1; end < n; ++end) { int cmp; if (arr[end - 1] == arr[end]) { cmp = 0; } else if (arr[end - 1] < arr[end]) { cmp = 1; } else { cmp = -1; } int nextCmp = 0; if (end < n - 1) { if (arr[end] == arr[end + 1]) { nextCmp = 0; } else if (arr[end] < arr[end + 1]) { nextCmp = 1; } else { nextCmp = -1; } } if (end == n - 1 || cmp * nextCmp != -1) { if (cmp != 0) { maxLen = Math.max(maxLen, end - start + 1); } start = end; } } return maxLen; } public static void main(String[] args) { // Example 1 int[] arr1 = {9, 4, 2, 10, 7, 8, 8, 1, 9}; System.out.println(maxTurbulenceSize(arr1)); // Output: 5 } } # code by flutterfly def max_turbulence_size(arr): n = len(arr) if n < 2: return n max_len = 1 start = 0 for end in range(1, n): if arr[end - 1] == arr[end]: cmp = 0 elif arr[end - 1] < arr[end]: cmp = 1 else: cmp = -1 if end == n - 1 or cmp * (0 if arr[end] == arr[end + 1] else (1 if arr[end] < arr[end + 1] else -1)) != -1: if cmp != 0: max_len = max(max_len, end - start + 1) start = end return max_len if __name__ == "__main__": # Example 1 arr1 = [9, 4, 2, 10, 7, 8, 8, 1, 9] print(max_turbulence_size(arr1)) # Output: 5
//code by flutterfly using System; class Program { public static int MaxTurbulenceSize(int[] arr) { int n = arr.Length; if (n < 2) return n; int maxLen = 1; int start = 0; for (int end = 1; end < n; ++end) { int cmp = arr[end - 1] == arr[end] ? 0 : (arr[end - 1] < arr[end] ? 1 : -1); if (end == n - 1 || cmp * (arr[end] == arr[end + 1] ? 0 : (arr[end] < arr[end + 1] ? 1 : -1)) != -1) { if (cmp != 0) { maxLen = Math.Max(maxLen, end - start + 1); } start = end; } } return maxLen; } static void Main() { // Example 1 int[] arr1 = { 9, 4, 2, 10, 7, 8, 8, 1, 9 }; Console.WriteLine(MaxTurbulenceSize(arr1)); // Output: 5 } } //code by flutterfly function maxTurbulenceSize(arr) { const n = arr.length; if (n < 2) return n; let maxLen = 1; let start = 0; for (let end = 1; end < n; ++end) { let cmp = arr[end - 1] === arr[end] ? 0 : (arr[end - 1] < arr[end] ? 1 : -1); if (end === n - 1 || cmp * (arr[end] === arr[end + 1] ? 0 : (arr[end] < arr[end + 1] ? 1 : -1)) !== -1) { if (cmp !== 0) { maxLen = Math.max(maxLen, end - start + 1); } start = end; } } return maxLen; } // Example 1 const arr1 = [9, 4, 2, 10, 7, 8, 8, 1, 9]; console.log(maxTurbulenceSize(arr1)); // Output: 5 Time Complexity: O(n), where n is the length of the array. This is because we are iterating through the array once.
Auxiliary Space: O(1) since we are using only a few extra variables to keep track of indices and the maximum length.
Similar Reads
Longest Subarray with 0 Sum Given an array arr[] of size n, the task is to find the length of the longest subarray with sum equal to 0.Examples:Input: arr[] = {15, -2, 2, -8, 1, 7, 10, 23}Output: 5Explanation: The longest subarray with sum equals to 0 is {-2, 2, -8, 1, 7}Input: arr[] = {1, 2, 3}Output: 0Explanation: There is n
10 min read
Longest increasing subarray Given an array containing n numbers. The problem is to find the length of the longest contiguous subarray such that every element in the subarray is strictly greater than its previous element in the same subarray. Time Complexity should be O(n). Examples: Input : arr[] = {5, 6, 3, 5, 7, 8, 9, 1, 2}
15+ min read
Longest common subarray in the given two arrays Given two arrays A[] and B[] of N and M integers respectively, the task is to find the maximum length of an equal subarray or the longest common subarray between the two given array. Examples: Input: A[] = {1, 2, 8, 2, 1}, B[] = {8, 2, 1, 4, 7} Output: 3 Explanation: The subarray that is common to b
15+ min read
Length of the longest alternating subarray Given an array of N including positive and negative numbers only. The task is to find the length of the longest alternating (means negative-positive-negative or positive-negative-positive) subarray present in the array. Examples: Input: a[] = {-5, -1, -1, 2, -2, -3} Output: 3 The subarray {-1, 2, -2
5 min read
Longest subarray having maximum sum Given an array arr[] containing n integers. The problem is to find the length of the subarray having maximum sum. If there exists two or more subarrays with maximum sum then print the length of the longest subarray.Examples: Input : arr[] = {5, -2, -1, 3, -4}Output : 4There are two subarrays with ma
12 min read