# 2457. Minimum Addition to Make Integer Beautiful

## Description

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.

Example 1:

Input: n = 16, target = 6
Output: 4
Explanation: Initially n is 16 and its digit sum is 1 + 6 = 7. After adding 4, n becomes 20 and digit sum becomes 2 + 0 = 2. It can be shown that we can not make n beautiful with adding non-negative integer less than 4.


Example 2:

Input: n = 467, target = 6
Output: 33
Explanation: Initially n is 467 and its digit sum is 4 + 6 + 7 = 17. After adding 33, n becomes 500 and digit sum becomes 5 + 0 + 0 = 5. It can be shown that we can not make n beautiful with adding non-negative integer less than 33.


Example 3:

Input: n = 1, target = 1
Output: 0
Explanation: Initially n is 1 and its digit sum is 1, which is already smaller than or equal to target.


Constraints:

• 1 <= n <= 1012
• 1 <= target <= 150
• The input will be generated such that it is always possible to make n beautiful.

## Solutions

Solution 1: Greedy Algorithm

We define a function $f(x)$ to represent the sum of the digits of an integer $x$. The problem is to find the minimum non-negative integer $x$ such that $f(n + x) \leq target$.

If the sum of the digits of $y = n+x$ is greater than $target$, we can loop through the following operations to reduce the sum of the digits of $y$ to less than or equal to $target$:

• Find the lowest non-zero digit of $y$, reduce it to $0$, and add $1$ to the digit one place higher;
• Update $x$ and continue the above operation until the sum of the digits of $n+x$ is less than or equal to $target$.

After the loop ends, return $x$.

For example, if $n=467$ and $target=6$, the change process of $n$ is as follows:

\begin{aligned} & 467 \rightarrow 470 \rightarrow 500 \\ \end{aligned}

The time complexity is $O(\log^2 n)$, where $n$ is the integer given in the problem. The space complexity is $O(1)$.

• class Solution {
public long makeIntegerBeautiful(long n, int target) {
long x = 0;
while (f(n + x) > target) {
long y = n + x;
long p = 10;
while (y % 10 == 0) {
y /= 10;
p *= 10;
}
x = (y / 10 + 1) * p - n;
}
return x;
}

private int f(long x) {
int y = 0;
while (x > 0) {
y += x % 10;
x /= 10;
}
return y;
}
}

• class Solution {
public:
long long makeIntegerBeautiful(long long n, int target) {
using ll = long long;
auto f = [](ll x) {
int y = 0;
while (x) {
y += x % 10;
x /= 10;
}
return y;
};

ll x = 0;
while (f(n + x) > target) {
ll y = n + x;
ll p = 10;
while (y % 10 == 0) {
y /= 10;
p *= 10;
}
x = (y / 10 + 1) * p - n;
}
return x;
}
};

• class Solution:
def makeIntegerBeautiful(self, n: int, target: int) -> int:
def f(x: int) -> int:
y = 0
while x:
y += x % 10
x //= 10
return y

x = 0
while f(n + x) > target:
y = n + x
p = 10
while y % 10 == 0:
y //= 10
p *= 10
x = (y // 10 + 1) * p - n
return x


• func makeIntegerBeautiful(n int64, target int) (x int64) {
f := func(x int64) (y int) {
for ; x > 0; x /= 10 {
y += int(x % 10)
}
return
}
for f(n+x) > target {
y := n + x
var p int64 = 10
for y%10 == 0 {
y /= 10
p *= 10
}
x = (y/10+1)*p - n
}
return
}

• function makeIntegerBeautiful(n: number, target: number): number {
const f = (x: number): number => {
let y = 0;
for (; x > 0; x = Math.floor(x / 10)) {
y += x % 10;
}
return y;
};

let x = 0;
while (f(n + x) > target) {
let y = n + x;
let p = 10;
while (y % 10 === 0) {
y = Math.floor(y / 10);
p *= 10;
}
x = (Math.floor(y / 10) + 1) * p - n;
}
return x;
}