Generate a number such that the frequency of each digit is digit times the frequency in given number

Given a number N containing digits from 1 to 9 only. The task is to generate a new number using the number N such that the frequency of each digit in the new number is equal to the frequency of that digit in N multiplied by the digit itself.

Note: The digits in the new number must be in increasing order.

Examples:

Input : N = 312
Output : 122333
Explanation : The output contains digit 1 once, digit 2 twice and digit 3 thrice.

Input : N = 525
Output : 225555555555
Explanation : The output contains digit 2 twice and digit 5 ten times. 5 is ten times because its frequency is 2 in the given integer.



The idea is to store the count or the frequency of the digits in the given number N using a counting array or hash. Now, for each digit add it to the new number, K number of times where K is equal to its frequency in the counting array multiplied by the digit itself.

Below is the implementation of the above approach:

C++

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// CPP program to print a number such that the
//  frequency of each digit in the new number is 
// is equal to its frequency in the given number 
// multiplied by the digit itself.
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to print such a number
void printNumber(int n)
{
    // initializing a hash array
    int count[10] = { 0 };
  
    // counting frequency of the digits
    while (n) {
        count[n % 10]++;
        n /= 10;
    }
  
    // printing the new number
    for (int i = 1; i < 10; i++) {
        for (int j = 0; j < count[i] * i; j++)
            cout << i;
    }
}
  
// Driver code
int main()
{
    int n = 3225;
  
    printNumber(n);
  
    return 0;
}

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Java














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// Java program to print a number such that the 


// frequency of each digit in the new number is  


// is equal to its frequency in the given number  


// multiplied by the digit itself. 


  


import java.io.*; 


  


class GFG { 


   


// Function to print such a number 


static void printNumber(int n) 


{ 


    // initializing a hash array 


    int count[] = new int[10]; 


  


    // counting frequency of the digits 


    while (n>0) { 


        count[n % 10]++; 


        n /= 10; 


    } 


  


    // printing the new number 


    for (int i = 1; i < 10; i++) { 


        for (int j = 0; j < count[i] * i; j++) 


            System.out.print(i); 


    } 


} 


  


// Driver code 


  


    public static void main (String[] args) { 


        int n = 3225; 


  


    printNumber(n); 


    } 


} 


// This code is contributed by inder_verma 



















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Python3














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# Python 3 program to print a number such that the 


# frequency of each digit in the new number is  


# is equal to its frequency in the given number  


# multiplied by the digit itself. 


  


# Function to print such a number 


def printNumber(n): 


  


    # initializing a hash array 


    count = [0]*10


  


    # counting frequency of the digits 


    while (n) : 


        count[n % 10] += 1


        n //= 10


  


    # printing the new number 


    for i in range(1,10) : 


        for j in range(count[i] * i): 


            print(i,end="") 


  


# Driver code 


if __name__ == "__main__": 


    n = 3225


  


    printNumber(n) 


      


# This code is contributed by 


# ChitraNayal 



















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C#














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// C# program to print a number such  


// that the frequency of each digit  


// in the new number is equal to its  


// frequency in the given number  


// multiplied by the digit itself. 


using System; 


  


class GFG  


{ 


  


// Function to print such a number 


static void printNumber(int n) 


{ 


    // initializing a hash array 


    int []count = new int[10]; 


  


    // counting frequency of  


    // the digits 


    while (n > 0)  


    { 


        count[n % 10]++; 


        n /= 10; 


    } 


  


    // printing the new number 


    for (int i = 1; i < 10; i++) 


    { 


        for (int j = 0;  


                 j < count[i] * i; j++) 


            Console.Write(i); 


    } 


} 


  


// Driver code 


public static void Main ()  


{ 


    int n = 3225; 


  


    printNumber(n); 


} 


} 


  


// This code is contributed 


// by inder_verma 



















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PHP














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<?php 


// PHP program to print a number such  


// that the frequency of each digit  


// in the new number is equal to its 


// frequency in the given number  


// multiplied by the digit itself. 


  


// Function to print such a number 


function printNumber($n) 


{ 


    // initializing a hash array 


    $count = array(); 


    for($i = 0; $i<= 10; $i++) 


        $count[$i] = 0; 


          


    // counting frequency of the digits 


    while ($n)  


    { 


        $count[$n % 10]++; 


        $n /= 10; 


    } 


  


    // printing the new number 


    for ($i = 1; $i < 10; $i++)  


    { 


        for ($j = 0;  


             $j < $count[$i] * $i; $j++) 


            echo $i; 


    } 


} 


  


// Driver code 


$n = 3225; 


  


printNumber($n); 


  


// This code is contributed  


// by Akanksha Rai(Abby_akku) 



















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Output:


222233355555





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