How do you solve Longest Turbulent Subarray on LeetCode?

Longest Turbulent Subarray (LeetCode #978) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: The pattern of comparisons is crucial for identifying turbulent subarrays.

What is the brute force solution for Longest Turbulent Subarray?

The brute force approach for Longest Turbulent Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks every possible subarray to see if it is turbulent. By comparing adjacent elements, we can determine if the subarray meets the turbulent criteria. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Longest Turbulent Subarray?

Longest Turbulent Subarray uses the following concepts: Array, Dynamic Programming, Sliding Window. The recommended patterns to study are: Sliding Window, Two Pointers.

What are the interview tips for Longest Turbulent Subarray?

Always consider edge cases first — they can help clarify your approach. Explain your thought process clearly as you code; it shows your understanding. Practice recognizing patterns in problems, as they can lead to more efficient solutions.

What are common mistakes when solving Longest Turbulent Subarray?

Not resetting the current length correctly when a non-turbulent pair is found. Overlooking edge cases like arrays with only one element.

#978

Longest Turbulent Subarray

Medium

ArrayDynamic ProgrammingSliding WindowSliding WindowTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal approach uses a single pass through the array to track the length of the current turbulent subarray. This is more efficient because we only need to check adjacent elements, rather than all possible subarrays.

⚙️

Algorithm

4 steps

1Step 1: Initialize two variables to track the current length of turbulent subarray and the maximum length found.
2Step 2: Iterate through the array, comparing adjacent elements to determine if they form a turbulent pair.
3Step 3: If they do, increase the current length. If not, reset the current length to 1.
4Step 4: Update the maximum length whenever the current length exceeds it.

solution.py16 lines

1# Full working Python code
2
3def maxTurbulenceSize(arr):
4    n = len(arr)
5    if n < 2:
6        return n
7    max_length = 1
8    current_length = 1
9    for i in range(1, n):
10        if (arr[i - 1] < arr[i] and (i % 2 == 1)) or (arr[i - 1] > arr[i] and (i % 2 == 0)):
11            current_length += 1
12        else:
13            current_length = 2 if arr[i - 1] != arr[i] else 1
14        max_length = max(max_length, current_length)
15    return max_length
16

ℹ

Complexity note: The time complexity is O(n) because we only traverse the array once. The space complexity is O(1) since we are using a constant amount of extra space.

1The pattern of comparisons is crucial for identifying turbulent subarrays.
2Tracking the length of the current turbulent subarray allows for efficient updates.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.