728. Self Dividing Numbers LeetCode Solution

In this guide, you will get 728. Self Dividing Numbers LeetCode Solution with the best time and space complexity. The solution to Self Dividing Numbers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Self Dividing Numbers solution in C++
Self Dividing Numbers solution in Java
Self Dividing Numbers solution in Python
Additional Resources

728. Self Dividing Numbers LeetCode Solution image

Problem Statement of Self Dividing Numbers

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.

A self-dividing number is not allowed to contain the digit zero.
Given two integers left and right, return a list of all the self-dividing numbers in the range [left, right] (both inclusive).

Example 1:
Input: left = 1, right = 22
Output: [1,2,3,4,5,6,7,8,9,11,12,15,22]
Example 2:
Input: left = 47, right = 85
Output: [48,55,66,77]

Constraints:

1 <= left <= right <= 104

Complexity Analysis

Time Complexity: O(n\log n)
Space Complexity: O(1)

728. Self Dividing Numbers LeetCode Solution in C++

class Solution {
 public:
  vector<int> selfDividingNumbers(int left, int right) {
    vector<int> ans;

    for (int num = left; num <= right; ++num)
      if (selfDividingNumbers(num))
        ans.push_back(num);

    return ans;
  }

 private:
  bool selfDividingNumbers(int num) {
    for (int n = num; n > 0; n /= 10)
      if (n % 10 == 0 || num % (n % 10) != 0)
        return false;
    return true;
  }
};
/* code provided by PROGIEZ */

728. Self Dividing Numbers LeetCode Solution in Java

class Solution {
  public List<Integer> selfDividingNumbers(int left, int right) {
    List<Integer> ans = new ArrayList<>();

    for (int num = left; num <= right; ++num)
      if (dividingNumber(num))
        ans.add(num);

    return ans;
  }

  private boolean dividingNumber(int num) {
    for (int n = num; n > 0; n /= 10)
      if (n % 10 == 0 || num % (n % 10) != 0)
        return false;
    return true;
  }
}
// code provided by PROGIEZ

728. Self Dividing Numbers LeetCode Solution in Python

class Solution:
  def selfDividingNumbers(self, left: int, right: int) -> list[int]:
    return [num for num in range(left, right + 1) if all(n != 0 and num % n == 0 for n in map(int, str(num)))]
 # code by PROGIEZ