#2765

Easy

LeetCode Problem Workspace

Longest Alternating Subarray

Find the longest alternating subarray in a given array of integers.

Problem Statement

You are given a 0-indexed integer array nums. A subarray s of length m is called alternating if for every consecutive pair of indices i and j in s, nums[i] != nums[i+1]. A subarray is a contiguous sequence of elements from the array.

Return the maximum length of all alternating subarrays present in nums or -1 if no such subarray exists. The array length is between 2 and 100, and each element is between 1 and 10^4.

Examples

Example 1

Input: nums = [2,3,4,3,4]

Output: 4

The alternating subarrays are [2, 3] , [3,4] , [3,4,3] , and [3,4,3,4] . The longest of these is [3,4,3,4] , which is of length 4.

Example 2

Input: nums = [4,5,6]

Output: 2

[4,5] and [5,6] are the only two alternating subarrays. They are both of length 2.

Constraints

2 <= nums.length <= 100
1 <= nums[i] <= 104

Solution Approach

Iterate through the array

Go through each element in the array, checking if consecutive elements alternate. Keep track of the longest alternating subarray encountered.

Use enumeration

By checking each possible subarray starting at each index, we can easily identify alternating subarrays and find the one with the maximum length.

Edge cases

Handle the case where no alternating subarray exists by returning -1. Consider cases where all elements are the same or there are no alternations.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

Time complexity can be O(n) where n is the length of the array, depending on the approach used to detect and count alternating subarrays. Space complexity is O(1) since we are not using extra space aside from counters and temporary variables.

What Interviewers Usually Probe

Candidate understands the pattern of alternating subarrays.
The candidate can efficiently loop through the array without unnecessary complexity.
Candidate considers edge cases such as no valid alternating subarrays.

Common Pitfalls or Variants

Common pitfalls

Not handling the case where no alternating subarrays exist, leading to an incorrect result.
Overcomplicating the solution with extra loops or unnecessary calculations.
Failing to properly track the maximum length of alternating subarrays.

Follow-up variants

Consider optimizing the approach for larger arrays.
Extend the problem to find alternating subarrays with specific length constraints.
Alter the problem to return the longest alternating subarray itself, not just the length.

FAQ

How do I find the longest alternating subarray in the problem "Longest Alternating Subarray"?

Iterate through the array, checking consecutive elements for alternation. Keep track of the longest subarray you find.

What is the time complexity of solving the "Longest Alternating Subarray" problem?

The time complexity is O(n) where n is the length of the array, assuming you efficiently check consecutive elements.

What should I do if no alternating subarray exists?

Return -1 if no alternating subarray is found in the array.

How do I handle edge cases for the "Longest Alternating Subarray" problem?

Consider arrays where no alternations occur, such as arrays with all equal elements. Also handle the minimum array length.

Can the "Longest Alternating Subarray" problem be solved by brute force?

While brute force is possible, it is not efficient. Use enumeration and careful iteration to optimize the solution.

terminal

Solution

Solution 1: Enumeration

We can enumerate the left endpoint $i$ of the subarray, and for each $i$, we need to find the longest subarray that satisfies the condition. We can start traversing to the right from $i$, and each time we encounter adjacent elements whose difference does not satisfy the alternating condition, we find a subarray that satisfies the condition. We can use a variable $k$ to record whether the difference of the current element should be $1$ or $-1$. If the difference of the current element should be $-k$, then we take the opposite of $k$. When we find a subarray $nums[i..j]$ that satisfies the condition, we update the answer to $\max(ans, j - i + 1)$.

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class Solution:
    def alternatingSubarray(self, nums: List[int]) -> int:
        ans, n = -1, len(nums)
        for i in range(n):
            k = 1
            j = i
            while j + 1 < n and nums[j + 1] - nums[j] == k:
                j += 1
                k *= -1
            if j - i + 1 > 1:
                ans = max(ans, j - i + 1)
        return ans

Continue Practicing

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