Introduction

In this blog post, we will explore how to solve the LeetCode problem "2310" 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 two integers num and k, consider a set of positive integers with the following properties: The units digit of each integer is k. The sum of the integers is num. Return the minimum possible size of such a set, or -1 if no such set exists. Note: The set can contain multiple instances of the same integer, and the sum of an empty set is considered 0. The units digit of a number is the rightmost digit of the number. Example 1: Input: num = 58, k = 9 Output: 2 Explanation: One valid set is [9,49], as the sum is 58 and each integer has a units digit of 9. Another valid set is [19,39]. It can be shown that 2 is the minimum possible size of a valid set. Example 2: Input: num = 37, k = 2 Output: -1 Explanation: It is not possible to obtain a sum of 37 using only integers that have a units digit of 2. Example 3: Input: num = 0, k = 7 Output: 0 Explanation: The sum of an empty set is considered 0. Constraints: 0 <= num <= 3000 0 <= k <= 9

Explanation

Here's the solution approach, complexity analysis, and Python code:

• Check for Base Cases: Handle the cases where num is 0 (return 0) or where it's fundamentally impossible to achieve the target num with numbers ending in k.
• Iterate and Check: Iterate from 1 up to 10 (since any solution > 10 can be reduced to <= 10). For each possible size i, check if num - (i * k) is non-negative and divisible by 10. If so, we've found a valid set size and return i.

• Handle No Solution: If the loop completes without finding a solution, it means no set exists. Return -1.

• Time Complexity: O(1). The loop runs for a maximum of 10 iterations, making it a constant time operation.

• Space Complexity: O(1). The algorithm uses only a constant amount of extra space.

Code

    def minimumNumbers(num: int, k: int) -> int:
    if num == 0:
        return 0

    for i in range(1, 11):
        if (num - (i * k)) >= 0 and (num - (i * k)) % 10 == 0:
            return i

    return -1