# 2310. Sum of Numbers With Units Digit K

## Description

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

## Solutions

• class Solution {
public int minimumNumbers(int num, int k) {
if (num == 0) {
return 0;
}
for (int i = 1; i <= num; ++i) {
int t = num - k * i;
if (t >= 0 && t % 10 == 0) {
return i;
}
}
return -1;
}
}

• class Solution {
public:
int minimumNumbers(int num, int k) {
if (num == 0) return 0;
for (int i = 1; i <= num; ++i) {
int t = num - k * i;
if (t >= 0 && t % 10 == 0) return i;
}
return -1;
}
};

• class Solution:
def minimumNumbers(self, num: int, k: int) -> int:
if num == 0:
return 0
for i in range(1, num + 1):
if (t := num - k * i) >= 0 and t % 10 == 0:
return i
return -1


• func minimumNumbers(num int, k int) int {
if num == 0 {
return 0
}
for i := 1; i <= num; i++ {
t := num - k*i
if t >= 0 && t%10 == 0 {
return i
}
}
return -1
}

• function minimumNumbers(num: number, k: number): number {
if (!num) return 0;
let digit = num % 10;
for (let i = 1; i < 11; i++) {
let target = i * k;
if (target <= num && target % 10 == digit) return i;
}
return -1;
}