Introduction

In this blog post, we will explore how to solve the LeetCode problem "413" 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

An integer array is called arithmetic if it consists of at least three elements and if the difference between any two consecutive elements is the same. For example, [1,3,5,7,9], [7,7,7,7], and [3,-1,-5,-9] are arithmetic sequences. Given an integer array nums, return the number of arithmetic subarrays of nums. A subarray is a contiguous subsequence of the array. Example 1: Input: nums = [1,2,3,4] Output: 3 Explanation: We have 3 arithmetic slices in nums: [1, 2, 3], [2, 3, 4] and [1,2,3,4] itself. Example 2: Input: nums = [1] Output: 0 Constraints: 1 <= nums.length <= 5000 -1000 <= nums[i] <= 1000

Explanation

Here's an efficient solution to count arithmetic subarrays:

• Dynamic Programming: We'll use dynamic programming to efficiently count arithmetic subarrays. We maintain a variable to store the count of arithmetic subarrays ending at the current index, reusing the previous count when the current element extends an existing arithmetic sequence.

• Iterative Approach: Iterate through the array, checking if the current element extends an existing arithmetic sequence. If it does, increment the count of arithmetic subarrays.

• Runtime Complexity: O(n), Storage Complexity: O(1)

Code

    def numberOfArithmeticSlices(nums):
    n = len(nums)
    if n < 3:
        return 0

    count = 0
    dp = 0  # dp[i] stores the number of arithmetic slices ending at index i

    for i in range(2, n):
        if nums[i] - nums[i - 1] == nums[i - 1] - nums[i - 2]:
            dp = dp + 1
            count = count + dp
        else:
            dp = 0  # Reset if the current element doesn't extend the sequence

    return count