How do you solve Longest Alternating Subarray on LeetCode?

Longest Alternating Subarray (LeetCode #2765) 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 alternating pattern can be identified by checking the difference between adjacent elements.

What is the brute force solution for Longest Alternating Subarray?

The brute force approach for Longest Alternating Subarray is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks all possible subarrays to see if they meet the alternating condition. This is straightforward but inefficient because it examines every possible combination. The optimal Optimal Solution approach improves this to O(n).

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

Longest Alternating Subarray uses the following concepts: Array, Enumeration. The recommended patterns to study are: Array, Two Pointers.

What are the interview tips for Longest Alternating Subarray?

Always clarify the problem statement and constraints before jumping into coding. Think about edge cases, such as arrays with no valid alternating subarrays. Start with a brute force approach to build intuition before optimizing.

What are common mistakes when solving Longest Alternating Subarray?

Not considering the case where the maximum length is less than 2 and returning -1. Failing to reset the current length when the alternating condition breaks.

#2765

Longest Alternating Subarray

Easy

ArrayEnumerationArrayTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach leverages a single pass through the array to count the length of valid alternating subarrays. This is efficient because we only need to check adjacent elements once.

⚙️

Algorithm

5 steps

1Step 1: Initialize a variable to keep track of the current length of the alternating subarray and a maximum length variable.
2Step 2: Loop through the array starting from the second element.
3Step 3: For each element, check if it alternates with the previous one. If it does, increment the current length; otherwise, reset it.
4Step 4: Update the maximum length if the current length exceeds it.
5Step 5: Return the maximum length found, or -1 if no valid subarray exists.

solution.py12 lines

1def longestAlternatingSubarray(nums):
2    max_length = -1
3    current_length = 0
4    n = len(nums)
5    for i in range(1, n):
6        if nums[i] == nums[i - 1] + 1 or nums[i] == nums[i - 1] - 1:
7            current_length += 1
8            if current_length > 0:
9                max_length = max(max_length, current_length + 1)
10        else:
11            current_length = 0
12    return max_length if max_length > 1 else -1

ℹ

Complexity note: The time complexity is O(n) because we only pass through the array once. The space complexity is O(1) since we only use a few variables for tracking lengths.

1The alternating pattern can be identified by checking the difference between adjacent elements.
2The problem can be solved efficiently with a single pass through the array.

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