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

Solution 1

Python3JavaC++GoTypeScript

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

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;
    }
};

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;
}

Solution 2

Python3JavaC++Go

class Solution:
    def minimumNumbers(self, num: int, k: int) -> int:
        if num == 0:
            return 0
        for i in range(1, 11):
            if (k * i) % 10 == num % 10 and k * i <= num:
                return i
        return -1

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

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

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

Solution 3

Python3

class Solution:
    def minimumNumbers(self, num: int, k: int) -> int:
        @cache
        def dfs(v):
            if v == 0:
                return 0
            if v < 10 and v % k:
                return inf
            i = 0
            t = inf
            while (x := i * 10 + k) <= v:
                t = min(t, dfs(v - x))
                i += 1
            return t + 1

        if num == 0:
            return 0
        if k == 0:
            return -1 if num % 10 else 1
        ans = dfs(num)
        return -1 if ans >= inf else ans

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