# 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.

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `using` `namespace` `std; ` ` `  `// Function to print such a number ` `void` `printNumber(``int` `n) ` `{ ` `    ``// initializing a hash array ` `    ``int` `count = { 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``; ` ` `  `    ``// 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

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Output:

```222233355555
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