Introduction

In this blog post, we will explore how to solve the LeetCode problem "2338" 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 two integers n and maxValue, which are used to describe an ideal array. A 0-indexed integer array arr of length n is considered ideal if the following conditions hold: Every arr[i] is a value from 1 to maxValue, for 0 <= i < n. Every arr[i] is divisible by arr[i - 1], for 0 < i < n. Return the number of distinct ideal arrays of length n. Since the answer may be very large, return it modulo 109 + 7. Example 1: Input: n = 2, maxValue = 5 Output: 10 Explanation: The following are the possible ideal arrays: - Arrays starting with the value 1 (5 arrays): [1,1], [1,2], [1,3], [1,4], [1,5] - Arrays starting with the value 2 (2 arrays): [2,2], [2,4] - Arrays starting with the value 3 (1 array): [3,3] - Arrays starting with the value 4 (1 array): [4,4] - Arrays starting with the value 5 (1 array): [5,5] There are a total of 5 + 2 + 1 + 1 + 1 = 10 distinct ideal arrays. Example 2: Input: n = 5, maxValue = 3 Output: 11 Explanation: The following are the possible ideal arrays: - Arrays starting with the value 1 (9 arrays): - With no other distinct values (1 array): [1,1,1,1,1] - With 2nd distinct value 2 (4 arrays): [1,1,1,1,2], [1,1,1,2,2], [1,1,2,2,2], [1,2,2,2,2] - With 2nd distinct value 3 (4 arrays): [1,1,1,1,3], [1,1,1,3,3], [1,1,3,3,3], [1,3,3,3,3] - Arrays starting with the value 2 (1 array): [2,2,2,2,2] - Arrays starting with the value 3 (1 array): [3,3,3,3,3] There are a total of 9 + 1 + 1 = 11 distinct ideal arrays. Constraints: 2 <= n <= 104 1 <= maxValue <= 104

Explanation

Here's a breakdown of the approach and the Python code:

• Dynamic Programming: Use dynamic programming to store the number of ideal arrays of length k ending with value v. The base case is when k = 1, where the number of such arrays is 1 for each value from 1 to maxValue.

• Iterative Calculation: Iterate through the possible lengths from 2 to n. For each length k and possible ending value v, calculate the number of ideal arrays by summing the number of ideal arrays of length k-1 ending with values that are divisors of v.

• Modulo Arithmetic: Apply modulo arithmetic to prevent integer overflow.

• Runtime & Storage Complexity: O(n maxValue log(maxValue)), due to the nested loops and divisor calculation. Space complexity is O(maxValue) for storing the DP table.

Code

    def idealArrays(n: int, maxValue: int) -> int:
    MOD = 10**9 + 7
    dp = [0] * (maxValue + 1)
    for i in range(1, maxValue + 1):
        dp[i] = 1

    for _ in range(2, n + 1):
        new_dp = [0] * (maxValue + 1)
        for i in range(1, maxValue + 1):
            for j in range(i, maxValue + 1, i):
                new_dp[j] = (new_dp[j] + dp[i]) % MOD
        dp = new_dp

    result = sum(dp) % MOD
    return result