Introduction

In this blog post, we will explore how to solve the LeetCode problem "446" 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 integer array nums, return the number of all the arithmetic subsequences of nums. A sequence of numbers 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. For example, [1, 1, 2, 5, 7] is not an arithmetic sequence. A subsequence of an array is a sequence that can be formed by removing some elements (possibly none) of the array. For example, [2,5,10] is a subsequence of [1,2,1,2,4,1,5,10]. The test cases are generated so that the answer fits in 32-bit integer. Example 1: Input: nums = [2,4,6,8,10] Output: 7 Explanation: All arithmetic subsequence slices are: [2,4,6] [4,6,8] [6,8,10] [2,4,6,8] [4,6,8,10] [2,4,6,8,10] [2,6,10] Example 2: Input: nums = [7,7,7,7,7] Output: 16 Explanation: Any subsequence of this array is arithmetic. Constraints: 1 <= nums.length <= 1000 -231 <= nums[i] <= 231 - 1

Explanation

Here's the breakdown of the solution:

• Dynamic Programming: Use a list of dictionaries (dp). Each dp[i] stores the count of arithmetic subsequences ending at index i with a specific difference. The key of the dictionary is the difference, and the value is the count.
• Iterate and Accumulate: Iterate through the nums array. For each number, iterate through the previous numbers. Calculate the difference between the current number and the previous number. Update the count of arithmetic subsequences ending at the current number, considering subsequences ending at the previous number and also starting new ones of length 3 or more.

• Total Count: Sum up all the counts of arithmetic subsequences to get the final result.

• Time Complexity: O(n^2), where n is the length of the input array. Space Complexity: O(n^2) in the worst case where all differences are unique, although often closer to O(n*k) where k is the number of unique differences.

Code

    def numberOfArithmeticSlices(nums):
    n = len(nums)
    dp = [{} for _ in range(n)]
    ans = 0

    for i in range(n):
        for j in range(i):
            diff = nums[i] - nums[j]

            if diff in dp[j]:
                count = dp[j][diff]
            else:
                count = 0

            if diff in dp[i]:
                dp[i][diff] += count + 1
            else:
                dp[i][diff] = count + 1

            ans += count

    return ans