# 3099. Harshad Number

## Description

An integer divisible by the sum of its digits is said to be a Harshad number. You are given an integer x. Return the sum of the digits of x if x is a Harshad number, otherwise, return -1.

Example 1:

Input: x = 18

Output: 9

Explanation:

The sum of digits of x is 9. 18 is divisible by 9. So 18 is a Harshad number and the answer is 9.

Example 2:

Input: x = 23

Output: -1

Explanation:

The sum of digits of x is 5. 23 is not divisible by 5. So 23 is not a Harshad number and the answer is -1.

Constraints:

• 1 <= x <= 100

## Solutions

### Solution 1: Simulation

We can calculate the sum of the digits of $x$, denoted as $s$, by simulation. If $x$ can be divided evenly by $s$, then we return $s$, otherwise, we return $-1$.

The time complexity is $O(\log x)$, where $x$ is the input integer. The space complexity is $O(1)$.

• class Solution {
public int sumOfTheDigitsOfHarshadNumber(int x) {
int s = 0;
for (int y = x; y > 0; y /= 10) {
s += y % 10;
}
return x % s == 0 ? s : -1;
}
}

• class Solution {
public:
int sumOfTheDigitsOfHarshadNumber(int x) {
int s = 0;
for (int y = x; y > 0; y /= 10) {
s += y % 10;
}
return x % s == 0 ? s : -1;
}
};

• class Solution:
def sumOfTheDigitsOfHarshadNumber(self, x: int) -> int:
s, y = 0, x
while y:
s += y % 10
y //= 10
return s if x % s == 0 else -1


• func sumOfTheDigitsOfHarshadNumber(x int) int {
s := 0
for y := x; y > 0; y /= 10 {
s += y % 10
}
if x%s == 0 {
return s
}
return -1
}

• function sumOfTheDigitsOfHarshadNumber(x: number): number {
let s = 0;
for (let y = x; y; y = Math.floor(y / 10)) {
s += y % 10;
}
return x % s === 0 ? s : -1;
}