# Solving Leetcode Interviews in Seconds with AI: Sum of Numbers With Units Digit K


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```