Sum of Numbers With Units Digit K - Problem

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.

Input & Output

Example 1 — Basic Case

$ Input: num = 58, k = 9

› Output: 2

💡 Note: We can use two numbers: 19 and 39. Both end in 9 and 19 + 39 = 58. This is the minimum possible size.

Example 2 — Single Number

$ Input: num = 37, k = 2

› Output: -1

💡 Note: No combination of positive numbers ending in 2 can sum to 37 (which ends in 7). Numbers ending in 2 create sums ending in even digits only.

Example 3 — Zero Sum

$ Input: num = 0, k = 7

› Output: 0

💡 Note: Empty set has sum 0, so we need 0 numbers.

Constraints

0 ≤ num ≤ 3000
0 ≤ k ≤ 9

Visualization

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Asked in

G Google 15 a Amazon 12 M Microsoft 8

The key insight is that the units digit pattern repeats every 10 numbers, so we only need to check counts 1-10. Best approach uses modular arithmetic for O(1) time complexity. Time: O(1), Space: O(1)

Common Approaches

✓ Brute Force Enumeration

⏱️ Time: O(num) Space: O(1)

For each possible count of numbers, check if we can form the target sum using numbers with units digit k. Start with the smallest positive number ending in k.

Mathematical Optimization

⏱️ Time: O(1) Space: O(1)

Instead of trying all counts, use the fact that if we have n numbers ending in k, their sum's units digit is (n*k)%10. We can solve for the minimum n directly.

Brute Force Enumeration — Algorithm Steps

Try count from 1 to num
For each count, check if (num - count*k) is divisible by 10
Return first valid count or -1

Visualization

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Step-by-Step Walkthrough

Try Count 1

Check if 1 number can sum to target

Try Count 2

Check if 2 numbers can sum to target

Found Valid

Return first valid count

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int solution(int num, int k) {
    if (num == 0) {
        return 0;
    }
    
    // Try each possible count of numbers
    for (int count = 1; count <= num; count++) {
        // Minimum sum with count numbers ending in k
        int minSum = count * k;
        if (minSum > num) {
            break;
        }
        
        // Check if we can reach num by adding multiples of 10
        int remaining = num - minSum;
        if (remaining >= 0 && remaining % 10 == 0) {
            return count;
        }
    }
    
    return -1;
}

int main() {
    int num, k;
    scanf("%d", &num);
    scanf("%d", &k);
    int result = solution(num, k);
    printf("%d\n", result);
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(num)

Loop through all possible counts from 1 to num

✓ Linear Growth

Space Complexity

O(1)

Only using constant variables

✓ Linear Space

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Sum of Numbers With Units Digit K - Problem

Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force Enumeration — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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