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A *self-dividing number* is a number that is divisible by every digit it contains.

For example, 128 is a self-dividing number because `128 % 1 == 0`

, `128 % 2 == 0`

, and `128 % 8 == 0`

.

Also, a self-dividing number is not allowed to contain the digit zero.

Given a lower and upper number bound, output a list of every possible self dividing number, including the bounds if possible.

**Example 1:**

Input:left = 1, right = 22Output:[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15, 22]

**Note:**

`1 <= left <= right <= 10000`

.b'

\n\n#### Approach #1: Brute Force [Accepted]

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**Intuition and Algorithm**

For each number in the given range, we will directly test if that number is self-dividing.

\nBy definition, we want to test each whether each digit is non-zero and divides the number. For example, with `128`

, we want to test `d != 0 && 128 % d == 0`

for `d = 1, 2, 8`

. To do that, we need to iterate over each digit of the number.

A straightforward approach to that problem would be to convert the number into a character array (string in Python), and then convert back to integer to perform the modulo operation when checking `n % d == 0`

.

We could also continually divide the number by 10 and peek at the last digit. That is shown as a variation in a comment.

\n\n**Complexity Analysis**

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Time Complexity: , where is the number of integers in the range , and assuming is bounded. (In general, the complexity would be .)

\n \n - \n
Space Complexity: , the length of the answer.

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Analysis written by: @awice.

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