Minimum Addition to Make Integer Beautiful

Minimum Addition to Make Integer Beautiful - Problem

Math Greedy Medium

You are given two positive integers n and target.

An integer is considered beautiful if the sum of its digits is less than or equal to target.

Return the minimum non-negative integer x such that n + x is beautiful. The input will be generated such that it is always possible to make n beautiful.

Input & Output

Example 1 — Basic Case

$ Input: n = 16, target = 6

› Output: 4

💡 Note: n=16 has digit sum 1+6=7 > 6. Adding 4 gives us 20 with digit sum 2+0=2 ≤ 6.

Example 2 — Already Beautiful

$ Input: n = 467, target = 20

› Output: 0

💡 Note: n=467 has digit sum 4+6+7=17 ≤ 20, so it's already beautiful.

Example 3 — Large Rounding

$ Input: n = 1, target = 1

› Output: 0

💡 Note: n=1 has digit sum 1 ≤ 1, so it's already beautiful.

Constraints

1 ≤ n ≤ 10¹²
1 ≤ target ≤ 150

Visualization

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Asked in

G Google 12 M Microsoft 8 a Amazon 5

The key insight is that rounding up digits to the next power of 10 can dramatically reduce digit sum. The greedy approach finds the optimal position to round up. Time: O(log n), Space: O(log n)

Common Approaches

✓ Brute Force

⏱️ Time: O(x × log(n+x)) Space: O(1)

Start with x=0 and keep incrementing x by 1. For each value, calculate the digit sum of n+x and check if it's ≤ target. Return the first x that works.

Greedy Digit Rounding

⏱️ Time: O(log n) Space: O(log n)

Convert number to string and process digits from right to left. When digit sum exceeds target, find the leftmost 'bad' digit and round it up to next power of 10, zeroing out all digits to its right.

Brute Force — Algorithm Steps

Start with x = 0
Calculate digit sum of n + x
If digit sum ≤ target, return x
Otherwise increment x and repeat

Visualization

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Step-by-Step Walkthrough

Start

Begin with x=0, check if n+0 is beautiful

Increment

If not beautiful, try x=1, x=2, x=3...

Found

Return first x where digit sum ≤ target

Code -

solution.c — C

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

int digitSum(long long num) {
    int total = 0;
    while (num > 0) {
        total += num % 10;
        num /= 10;
    }
    return total;
}

long long solution(long long n, int target) {
    long long x = 0;
    while (digitSum(n + x) > target) {
        x++;
    }
    return x;
}

int main() {
    long long n;
    int target;
    scanf("%lld", &n);
    scanf("%d", &target);
    printf("%lld\n", solution(n, target));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(x × log(n+x))

We may need to try up to x values, and calculating digit sum takes O(log(number)) time

⚡ Linearithmic

Space Complexity

O(1)

Only using a few variables for counting

✓ Linear Space

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Input & Output

Constraints

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Related Problems

Common Approaches

Brute Force — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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