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Find all the possible numbers in a range that can be evenly divided by its digits

  • Last Updated : 19 May, 2021

Given two integers n and m, which represent a range where n is the lower bound and m is the upper bound respectively. The task is to find all the possible numbers that lie between n and m that can be evenly divided by their digits.

Examples:

Input: n = 1, m = 15 
Output: [1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15] 
Explanation: 
The numbers in the array can be divided evenly by their digits except 
2 numbers 10 and 13, which lie between 1 and 15. 
10 can be broken into digits 1 and 0. 
10 % 1 == 0 but 10 % 0 != 0. 
13 can be broken into digits 1 and 3. 
13 % 1 == 0 but 13 % 3 != 0.

Input: n = 21, m = 25 
Output: [22, 24] 
Explanation: 
The numbers in the array can be divided evenly by their digits except 
there are 3 numbers; 21, 23 and 25 lie between 21 and 25. 
21 can be broken into digits 2 and 1. 
21 % 2 != 0 and 21 % 1 != 0 
23 can be broken into digits 2 and 3. 
23 % 2 != 0 and 23 % 3 != 0. 
25 can be broken into digits 2 and 5. 
25 % 2 != 0 but 25 % 5 == 0. 
 

Approach: 
The main idea is to iterate each and every number from left to right. For each number, convert it into a string followed by a character array and check whether it contains 0. If it contains 0, ignore it. Otherwise, for each number check, whether the number is evenly divided by its digits or not. If the number is evenly divided by its digits, then add it to the list, else discard it. Finally, return the list.



Below is the implementation of the above approach:  

C++




// C++ program to find all the
// possible numbers that can be
// evenly divided by its digits
#include<bits/stdc++.h>
#include<sstream>
#include<string>
using namespace std;
 
// Function to check whether the
// number is evenly divisible by
// its digits or not.
bool isDivisible(int num)
{
  // Iterate each number convert
  // number into string and then
  // to character array.
 
  // declaring output string stream
  ostringstream str1;
 
  // Sending a number as a stream
  // into output string
  str1 << num;
 
  // the str() coverts number into
  // string
  string str2 = str1.str();
 
  for (char c : str2 )
  {
    if (c == '0' ||
        num % (c - '0') > 0)
    {
      return false;
    }
  }
  return true;
}
 
// Function to check each and every
// number from left to right. If the
// number is divisible by its digits
// then the number is added into the list
vector<int> selfDividingNumber(int left,
                               int right)
{
  vector<int>list ;
   
  for (int i = left; i <= right; i++)
  {
    if (isDivisible(i))
    {
      list.push_back(i);
    }
  }
  return list;
}
 
// Driver Code
int main()
{
  // initialise range
  int n1 = 1, m1 = 15;
 
  vector<int> ans =
  (selfDividingNumber(n1, m1));
   
  for(auto i = ans.begin();
           i != ans.end(); i++)
    cout << (*i) << " ";
}
 
// This code is contributed by Chitranayal

Java




// Java program to find all the possible numbers
// that can be evenly divided by its digits
 
import java.util.*;
class GFG {
 
    // Function to check each and every number
    // from left to right. If the number is
    // divisible by its digits
    // then the number is added into the list
    static List<Integer> selfDividingNumber(int left,
                                            int right)
    {
 
        List<Integer> list = new ArrayList<Integer>();
        for (int i = left; i <= right; i++) {
            if (isDivisible(i)) {
                list.add(i);
            }
        }
        return list;
    }
 
    // Function to check whether the number
    // is evenly divisible by its digits or not.
    static boolean isDivisible(int num)
    {
 
        // Iterate each number convert number
        // into string and then to character array.
        for (char c : String.valueOf(num).toCharArray()) {
            if (c == '0' || num % (c - '0') > 0) {
                return false;
            }
        }
        return true;
    }
 
    // Driver Code
    public static void main(String args[])
    {
 
        // initialise range
        int n1 = 1, m1 = 15;
 
        System.out.print(selfDividingNumber(n1, m1));
    }
}

Python3




# Python3 program to find all the possible numbers
# that can be evenly divided by its digits
 
# Function to check each and every number
# from left to right. If the number is
# divisible by its digits
# then the number is added into the list
def selfDividingNumber(left, right) :
    array_list = [];
     
    for i in range(left, right + 1) :
        if (isDivisible(i)) :
            array_list.append(i);
             
    return array_list;
     
# Function to check whether the number
# is evenly divisible by its digits or not.
def isDivisible(num) :
     
    # Iterate each number convert number
    # into string and then to character array.
    for c in list(str(num)) :
        if (c == '0' or num % (ord(c) - ord('0')) > 0):
            return False;
             
    return True;
     
# Driver Code
if __name__ == "__main__" :
     
    # Initialise range
    n1 = 1; m1 = 15;
    print(selfDividingNumber(n1, m1));
 
# This code is contributed by AnkitRai01

C#




// C# program to find all the
// possible numbers that can
// be evenly divided by its digits
using System;
using System.Collections.Generic;
 
public class GFG {
 
// Function to check each and every number
// from left to right. If the number is
// divisible by its digits then the number
// is added into the list
static List<int> selfDividingNumber(int left,
                                    int right)
{
    List<int> list = new List<int>();
    for(int i = left; i <= right; i++)
    {
       if (isDivisible(i))
       {
           list.Add(i);
       }
    }
    return list;
}
 
// Function to check whether the number
// is evenly divisible by its digits or not.
static bool isDivisible(int num)
{
 
    // Iterate each number convert number
    // into string and then to character array.
    foreach(char c in String.Join("", num).ToCharArray())
    {
        if (c == '0' || num % (c - '0') > 0)
        {
            return false;
        }
    }
    return true;
}
 
// Driver Code
public static void Main(String []args)
{
 
    // Initialise range
    int n1 = 1, m1 = 15;
    List<int> t = selfDividingNumber(n1, m1);
     
    foreach(int val in t)
        Console.Write(val + ", ");
}
}
 
// This code is contributed by sapnasingh4991

Javascript




<script>
 
// Javascript program to find all the possible numbers
// that can be evenly divided by its digits
 
   // Function to check each and every number
    // from left to right. If the number is
    // divisible by its digits
    // then the number is added into the list
    function selfDividingNumber(left, right)
    {
  
        let list = [];
        for (let i = left; i <= right; i++) {
            if (isDivisible(i)) {
                list.push(i);
            }
        }
        return list;
    }
  
    // Function to check whether the number
    // is evenly divisible by its digits or not.
    function isDivisible(numm)
    {
         let num = numm.toString();
        // Iterate each number convert number
        // into string and then to character array.
        for (let c in num.split('')) {
            if (num == '0' || num % (num  - '0') > 0) {
                return false;
            }
        }
        return true;
    }
 
// Driver Code
     
    // initialise range
    let n1 = 1, m1 = 15;
  
    document.write(selfDividingNumber(n1, m1));
                       
</script>
Output: 
[1, 2, 3, 4, 5, 6, 7, 8, 9, 11, 12, 15]

 

Time Complexity: O(N), where N is the number of integers from left to right.
 

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