LeetCode-in-Java

3345. Smallest Divisible Digit Product I

Easy

You are given two integers n and t. Return the smallest number greater than or equal to n such that the product of its digits is divisible by t.

Example 1:

Input: n = 10, t = 2

Output: 10

Explanation:

The digit product of 10 is 0, which is divisible by 2, making it the smallest number greater than or equal to 10 that satisfies the condition.

Example 2:

Input: n = 15, t = 3

Output: 16

Explanation:

The digit product of 16 is 6, which is divisible by 3, making it the smallest number greater than or equal to 15 that satisfies the condition.

Constraints:

• 1 <= n <= 100
• 1 <= t <= 10

Solution

public class Solution {
    public int smallestNumber(int n, int t) {
        int num = -1;
        int check = n;
        while (num == -1) {
            int product = findProduct(check);
            if (product % t == 0) {
                num = check;
            }
            check += 1;
        }
        return num;
    }

    private int findProduct(int check) {
        int res = 1;
        while (check > 0) {
            res *= check % 10;
            check = check / 10;
        }
        return res;
    }
}

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