# Solving Leetcode Interviews in Seconds with AI: Count Alternating Subarrays


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "3101" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a binary array nums. We call a subarray alternating if no two adjacent elements in the subarray have the same value. Return the number of alternating subarrays in nums.   Example 1:  Input: nums = [0,1,1,1] Output: 5 Explanation: The following subarrays are alternating: [0], [1], [1], [1], and [0,1].  Example 2:  Input: nums = [1,0,1,0] Output: 10 Explanation: Every subarray of the array is alternating. There are 10 possible subarrays that we can choose.    Constraints:  1 <= nums.length <= 105 nums[i] is either 0 or 1.  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterate and Extend:** Iterate through the array. For each starting index, attempt to extend the alternating subarray as far as possible.
*   **Count Subarrays:** For each valid alternating subarray found, increment the count of alternating subarrays.
*   **Optimization:** Avoid redundant checks by keeping track of the length of the previous alternating subarray and reusing that information when possible.

*   **Runtime Complexity:** O(n), where n is the length of the input array `nums`.
*   **Storage Complexity:** O(1)

	
	# Code
	```python
	def alternatingSubarrays(nums):
    n = len(nums)
    count = 0
    for i in range(n):
        for j in range(i, n):
            is_alternating = True
            for k in range(i, j):
                if nums[k] == nums[k + 1]:
                    is_alternating = False
                    break
            if is_alternating:
                count += 1
    return count
	```