LeetCode 728 - Self Dividing Numbers

https://leetcode.com/problems/self-dividing-numbers/description/

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

Note:

The boundaries of each input argument are 1 <= left <= right <= 10000.

    public List<Integer> selfDividingNumbers(int left, int right) {
        List<Integer> ans = new ArrayList();
        for (int n = left; n <= right; ++n) {
            if (selfDividing(n)) ans.add(n);
        }
        return ans;
    }
    public boolean selfDividing(int n) {
        for (char c: String.valueOf(n).toCharArray()) {
            if (c == '0' || (n % (c - '0') > 0))
                return false;
        }
        return true;
    }
    /*
    Alternate implementation of selfDividing:
    public boolean selfDividing(int n) {
        int x = n;
        while (x > 0) {
            int d = x % 10;
            x /= 10;
            if (d == 0 || (n % d) > 0) return false;
        }
        return true;
    */

public List<Integer> selfDividingNumbers(int left, int right) {

  return IntStream.rangeClosed(left, right).boxed().filter(this::isSelfDividingNumber).collect(Collectors.toList());

}

private boolean isSelfDividingNumber(int num) {

  return !String.valueOf(num).chars().map(li -> li - '0').boxed().anyMatch(i -> i == 0 || num % i != 0);

  // Set<Integer> set = String.valueOf(num).chars().map(li -> li -

  // '0').boxed().collect(Collectors.toSet());

  //

  // for (int div : set) {

  // if (div == 0)

  // return false;

  // if (num % div != 0)

  // return false;

  // }

  //

  // return true;

}

https://leetcode.com/problems/self-dividing-numbers/discuss/109424/Java-solution-with-explanation-no-toString()-conversion

The idea is to traverse each integer sequence from left to right incrementing by one (left, left + 1, left + 2,..., left + n, right). Each time check if the the current number i is self-divided.
Self-division check is done by using '%' operator(we check each digit of i moving from right to left)
E.g. if i = 128 number:

1. 128 % 10 = 8, check 8 != 0;
2. remove 8 from next step j = 128 / 10 = 12
3. repeat 1 and 2 until j == 0

Time complexity is O(nm), where n = right - left, and m is number of digits in iterated number

    public List<Integer> selfDividingNumbers(int left, int right) {
        List<Integer> list = new ArrayList<>();
        for (int i = left; i <= right; i++) {
            int j = i;
            for (; j > 0; j /= 10) {
                if ((j % 10 == 0) || (i % (j % 10) != 0)) break;
            }
            if (j == 0) list.add(i); 
        }
        return list;
    }

https://leetcode.com/problems/self-dividing-numbers/discuss/109458/JavaC++-Clean-Code