README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/3000-3099/3099.Harshad%20Number/README_EN.md	1100	Weekly Contest 391 Q1	Math

3099. Harshad Number

中文文档

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 $O(1)$.

Python3

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

Java

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;
    }
}

C++

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;
    }
};

Go

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
}

TypeScript

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;
}