Numbers Divisible by Their Own Digits

Easy

Call a number *self-divisible* if it is evenly divisible by every one of its digits. For example, 36 is self-divisible: it divides by both 3 and 6. A number containing the digit 0 is never self-divisible, since you cannot divide by zero. Given two integers low and high, return all self-divisible numbers from low to high inclusive, in increasing order.

Example 1

low=10, high=30 → [11, 12, 15, 22, 24]

For example, 24 divides by both 2 and 4, while 20 is out because of its zero and 13 because it does not divide by 3.

Example 2

low=5, high=9 → [5, 6, 7, 8, 9]

A one-digit number always divides by itself, so all of them qualify.

Example 3

low=101, high=110 → []

Every number from 101 to 110 contains the digit 0, so the answer is empty.

Constraints

1 ≤ low ≤ high ≤ 10⁴

Have you seen this problem in a real interview?

Solved by0

Tried by0

Submissions0

Acceptance—

Topics

Math

Similar problems

A Number That Reads Both Ways Easy Value of a Roman Numeral Easy Add One to a Digit Array Easy Sum of Binary Strings Easy Whole Square Root Easy

Input

low=10, high=30

Expected

[11, 12, 15, 22, 24]

Run checks the first 3 tests in your browser. Submit checks all 81 tests on the server, hidden ones included.