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3622. Check Divisibility by Digit Sum and Product

Description

You are given a positive integer n. Determine whether n is divisible by the sum of the following two values:

  • The digit sum of n (the sum of its digits).

  • The digit product of n (the product of its digits).

Return true if n is divisible by this sum; otherwise, return false.

 

Example 1:

Input: n = 99

Output: true

Explanation:

Since 99 is divisible by the sum (9 + 9 = 18) plus product (9 * 9 = 81) of its digits (total 99), the output is true.

Example 2:

Input: n = 23

Output: false

Explanation:

Since 23 is not divisible by the sum (2 + 3 = 5) plus product (2 * 3 = 6) of its digits (total 11), the output is false.

 

Constraints:

  • 1 <= n <= 106

Solutions

Solution 1: Simulation

We can iterate through each digit of the integer \(n\), calculating the digit sum \(s\) and digit product \(p\). Finally, we check whether \(n\) is divisible by \(s + p\).

The time complexity is \(O(\log n)\), where \(n\) is the value of the integer \(n\). The space complexity is \(O(1)\).

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class Solution:
    def checkDivisibility(self, n: int) -> bool:
        s, p = 0, 1
        x = n
        while x:
            x, v = divmod(x, 10)
            s += v
            p *= v
        return n % (s + p) == 0

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class Solution {
    public boolean checkDivisibility(int n) {
        int s = 0, p = 1;
        int x = n;
        while (x != 0) {
            int v = x % 10;
            x /= 10;
            s += v;
            p *= v;
        }
        return n % (s + p) == 0;
    }
}

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class Solution {
public:
    bool checkDivisibility(int n) {
        int s = 0, p = 1;
        int x = n;
        while (x != 0) {
            int v = x % 10;
            x /= 10;
            s += v;
            p *= v;
        }
        return n % (s + p) == 0;
    }
};

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func checkDivisibility(n int) bool {
    s, p := 0, 1
    x := n
    for x != 0 {
        v := x % 10
        x /= 10
        s += v
        p *= v
    }
    return n%(s+p) == 0
}

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function checkDivisibility(n: number): boolean {
    let [s, p] = [0, 1];
    let x = n;
    while (x !== 0) {
        const v = x % 10;
        x = Math.floor(x / 10);
        s += v;
        p *= v;
    }
    return n % (s + p) === 0;
}

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