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++

// 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;
}

Java

// 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

Python3

# 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

C#

// 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

PHP

<?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)

Javascript

<script>

// JavaScript 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
function printNumber(n)
{
    // initializing a hash array
    let count = new Uint8Array(10);

    // counting frequency of the digits
    while (n) {
        count[n % 10]++;
        n = Math.floor(n / 10);
    }

    // printing the new number
    for (let i = 1; i < 10; i++) {
        for (let j = 0; j < count[i] * i; j++)
            document.write(i);
    }
}

// Driver code

    let n = 3225;

    printNumber(n);


// This code is contributed by Surbhi Tyagi.

</script>

Output:

222233355555

Time Complexity : O(log₁₀n), where n is the given integer.
Auxiliary Space : O(1), no extra space required so it is a constant.