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, 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

    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