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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 $O(1)$.

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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

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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;
    }
}

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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;
    }
};

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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
}

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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;
}

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