Longest Turbulent Array Leetcode Problem 978 [Python Solution]

In this blog post, we will tackle the LeetCode problem 978, "Longest Turbulent Subarray." This problem falls under the category of "Greedy" and is classified as a medium-level difficulty question.

We'll walk you through the problem, its constraints, and provide a Python solution to solve it efficiently.

Problem Overview

The problem statement is as follows: Given an integer array arr, your task is to return the length of the maximum size turbulent subarray of arr.

A subarray is considered turbulent if the comparison sign alternates between each adjacent pair of elements in the subarray.

Formally, a subarray [arr[i], arr[i + 1], ..., arr[j]] of arr is considered 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.

Let's take a look at a few examples to better understand this:

Example 1:

Input: arr = [9,4,2,10,7,8,8,1,9]

Output: 5

Explanation: In this case, the turbulent subarray is [2,10,7,8,8], where the comparison signs alternate correctly.

Example 2:

Input: arr = [4,8,12,16]

Output: 2

Example 3:

Input: arr = [100]

Output: 1

Constraints:

• 1 <= arr.length <= 4 * 10^4
• 0 <= arr[i] <= 10^9

Now that we've understood the problem statement and its requirements, let's delve into the solution.

Efficient Python Code Solution

We will implement a sliding window approach to efficiently solve this problem.

The idea is to maintain a sliding window, ensuring that the elements within the window form a valid turbulent subarray.

When the turbulent condition breaks, we will adjust the window accordingly to continue searching for the longest turbulent subarray.

Here's the Python solution:

def maxTurbulenceSize(arr):
    left, right = 0, 1  # Initialize left and right pointers for the sliding window.

result, prev = 1, ""  # Initialize the result and previous sign.

while right < len(arr):
        if arr[right - 1] > arr[right] and prev != ">":
            result = max(result, right - left + 1)
            right += 1
            prev = ">"
        elif arr[right - 1] < arr[right] and prev != "<":
            result = max(result, right - left + 1)
            right += 1
            prev = "<"
        else:
            right = right + 1 if arr[right] == arr[right - 1] else right
            left = right - 1
            prev = ""

    return result

This code efficiently solves the problem by maintaining a sliding window, identifying the longest turbulent subarray, and updating the result accordingly.

Time and Space Complexity

• Time Complexity: The time complexity of this solution is O(n), where n is the length of the input array arr.

We iterate through the array once, and the while loop inside the function ensures that each element is considered only once.

• Space Complexity: The space complexity is O(1), indicating that we do not use any additional data structures that depend on the input size.

The algorithm uses a constant amount of memory.

Related Interview Questions By Company:

• Two City Scheduling LeetCode
• Clone Graph LeetCode
• Unique Length 3 Palindromic Subsequences LeetCode

Related Interview Questions By Difficulty:

• Longest Turbulent Array LeetCode
• Maximum Sum Circular Subarray LeetCode
• Merge Triplets To Form Target Triplet LeetCode

Related Interview Questions By Category:

• Climbing Stairs LeetCode
• Combination Sum II LeetCode
• Validate Binary Search Tree LeetCode

Reasoning Behind Our Approach

The key to solving this problem efficiently lies in recognizing the alternating pattern required for a subarray to be turbulent.

By maintaining a sliding window and tracking the comparison signs, we can identify the longest turbulent subarray while minimizing redundant calculations.

The sliding window approach allows us to find the solution with a single pass through the array, resulting in a time-efficient algorithm.

In conclusion, we've successfully addressed the LeetCode problem 978, "Longest Turbulent Subarray," using a sliding window approach.

This approach optimally utilizes the given constraints and efficiently finds the length of the maximum turbulent subarray in the input array arr.

If you found this explanation and solution helpful, please like and engage to our content.

If you're preparing for coding interviews or seeking more coding resources, be sure to check out our website, Auditorical, where you'll find a wealth of free resources to help you prepare.

Thank you for reading, and we hope to see you again soon!

Question Link

If you have any questions, suggestions, or comments, please feel free to share them.

We encourage you to engage with the content and ask for any additional clarification or assistance you may need.

Happy coding!