# LC 2765. Longest Alternating Subarray

### [Problem link](https://leetcode.com/problems/longest-alternating-subarray/)

###### tags: `leedcode` `python` `easy`

You are given a **0-indexed** integer array <code>nums</code>. A subarray <code>s</code> of length <code>m</code> is called **alternating** if:

- <code>m</code> is greater than <code>1</code>.
- <code>s₁ = s₀ + 1</code>.
- The **0-indexed** subarray <code>s</code> looks like <code>[s₀, s₁, s₀, s₁,...,s_{(m-1) % 2}]</code>. In other words, <code>s₁ - s₀ = 1</code>, <code>s₂ - s₁ = -1</code>, <code>s₃ - s₂ = 1</code>, <code>s₄ - s₃ = -1</code>, and so on up to <code>s[m - 1] - s[m - 2] = (-1)^m</code>.

Return the maximum length of all **alternating** subarrays present in <code>nums</code> or <code>-1</code> if no such subarray exists.

A subarray is a contiguous **non-empty** sequence of elements within an array.

**Example 1:**

Input: nums = [2,3,4,3,4]
Output: 4
Explanation: The alternating subarrays are [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
Explanation: [4,5] and [5,6] are the only two alternating subarrays. They are both of length 2.

**Constraints:**

- <code>2 <= nums.length <= 100</code>
- <code>1 <= nums[i] <= 10⁴</code>

## Solution 1

class Solution:
    def alternatingSubarray(self, nums: List[int]) -> int:
        n = len(nums)
        res = -1
        
        for i in range(n):
            for j in range(i + 1, n):
                if nums[j] - nums[i] != (j - i) % 2:
                    break
                res = max(res, j - i + 1)
                
        return res

>### Complexity
>n = nums.length
>| | Time Complexity | Space Complexity |
>| ----------- | --------------- | ---------------- |
>| Solution 1 | O($n^2$) | O(1) |

## Note
x