Introduction

In this blog post, we will explore how to solve the LeetCode problem "1027" 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

Given an array nums of integers, return the length of the longest arithmetic subsequence in nums. Note that: A subsequence is an array that can be derived from another array by deleting some or no elements without changing the order of the remaining elements. A sequence seq is arithmetic if seq[i + 1] - seq[i] are all the same value (for 0 <= i < seq.length - 1). Example 1: Input: nums = [3,6,9,12] Output: 4 Explanation: The whole array is an arithmetic sequence with steps of length = 3. Example 2: Input: nums = [9,4,7,2,10] Output: 3 Explanation: The longest arithmetic subsequence is [4,7,10]. Example 3: Input: nums = [20,1,15,3,10,5,8] Output: 4 Explanation: The longest arithmetic subsequence is [20,15,10,5]. Constraints: 2 <= nums.length <= 1000 0 <= nums[i] <= 500

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: The core idea is to use dynamic programming to store the length of the longest arithmetic subsequence ending at each index with a specific common difference. We'll use a dictionary (hash map) to represent dp[i], where dp[i][diff] stores the length of the longest arithmetic subsequence ending at index i with common difference diff.
• Iterate and Update: Iterate through the nums array. For each element nums[i], iterate through the previous elements nums[j] (where j < i). Calculate the common difference diff = nums[i] - nums[j].

• Maximize Length: If the common difference has been seen before ending at index j, then the longest arithmetic subsequence ending at index i with difference diff is simply dp[j][diff] + 1. Otherwise, it's 2 (representing the subsequence nums[j], nums[i]). Update dp[i][diff] and keep track of the overall maximum length.

• Runtime & Storage Complexity: O(n^2), O(n^2)

Code

    def longestArithSeqLength(nums):
    n = len(nums)
    dp = [{} for _ in range(n)]
    max_len = 1

    for i in range(1, n):
        for j in range(i):
            diff = nums[i] - nums[j]
            if diff in dp[j]:
                dp[i][diff] = dp[j][diff] + 1
            else:
                dp[i][diff] = 2
            max_len = max(max_len, dp[i][diff])

    return max_len