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# 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.

Brute Force Approach:

We can use a map to store the frequency of each digit in the given number. Then, for each digit from 1 to 9, we can append that digit to the output string as many times as its frequency multiplied by the digit itself. Finally, we can sort the output string in increasing order and print it.

Below is the implementation of the above approach:

## C++

 `#include ``using` `namespace` `std;` `// Function to print such a number``void` `printNumber(``int` `n)``{``    ``// Count the frequency of each digit``    ``map<``int``, ``int``> freq;``    ``while` `(n > 0) {``        ``freq[n % 10]++;``        ``n /= 10;``    ``}` `    ``// Construct the output string``    ``string output = ``""``;``    ``for` `(``int` `i = 1; i <= 9; i++) {``        ``if` `(freq[i] > 0) {``            ``for` `(``int` `j = 0; j < freq[i] * i; j++)``                ``output += to_string(i);``        ``}``    ``}` `    ``// Sort and print the output string``    ``sort(output.begin(), output.end());``    ``cout << output;``}` `// Driver code``int` `main()``{``    ``int` `n = 3225;``    ``printNumber(n);``    ``return` `0;``}`

## Java

 `import` `java.util.*;` `public` `class` `Main``{` `  ``// Function to print such a number``  ``public` `static` `void` `printNumber(``int` `n)``  ``{` `    ``// Count the frequency of each digit``    ``Map freq = ``new` `HashMap();``    ``while` `(n > ``0``) {``      ``int` `digit = n % ``10``;``      ``freq.put(digit, freq.getOrDefault(digit, ``0``) + ``1``);``      ``n /= ``10``;``    ``}` `    ``// Construct the output string``    ``StringBuilder output = ``new` `StringBuilder();``    ``for` `(``int` `i = ``1``; i <= ``9``; i++) {``      ``if` `(freq.containsKey(i) && freq.get(i) > ``0``) {``        ``for` `(``int` `j = ``0``; j < freq.get(i) * i; j++)``          ``output.append(Integer.toString(i));``      ``}``    ``}` `    ``// Sort and print the output string``    ``char``[] arr = output.toString().toCharArray();``    ``Arrays.sort(arr);``    ``System.out.println(``new` `String(arr));``  ``}` `  ``// Driver code``  ``public` `static` `void` `main(String[] args) {``    ``int` `n = ``3225``;``    ``printNumber(n);``  ``}``}`

## Python3

 `# Function to print such a number``def` `printNumber(n):``    ``# Count the frequency of each digit``    ``freq ``=` `{}``    ``while` `n > ``0``:``        ``freq[n ``%` `10``] ``=` `freq.get(n ``%` `10``, ``0``) ``+` `1``        ``n ``/``/``=` `10` `    ``# Construct the output string``    ``output ``=` `""``    ``for` `i ``in` `range``(``1``, ``10``):``        ``if` `freq.get(i, ``0``) > ``0``:``            ``for` `j ``in` `range``(freq[i] ``*` `i):``                ``output ``+``=` `str``(i)` `    ``# Sort and print the output string``    ``print``(''.join(``sorted``(output)))`  `# Driver code``n ``=` `3225``printNumber(n)`

## C#

 `using` `System;``using` `System.Collections.Generic;``using` `System.Linq;` `class` `Program {` `    ``// Function to print such a number``    ``static` `void` `PrintNumber(``int` `n)``    ``{``        ``// Count the frequency of each digit``        ``Dictionary<``int``, ``int``> freq``            ``= ``new` `Dictionary<``int``, ``int``>();``        ``while` `(n > 0) {``            ``if` `(freq.ContainsKey(n % 10)) {``                ``freq[n % 10]++;``            ``}``            ``else` `{``                ``freq.Add(n % 10, 1);``            ``}``            ``n /= 10;``        ``}` `        ``// Construct the output string``        ``string` `output = ``""``;``        ``for` `(``int` `i = 1; i <= 9; i++) {``            ``if` `(freq.ContainsKey(i) && freq[i] > 0) {``                ``for` `(``int` `j = 0; j < freq[i] * i; j++) {``                    ``output += i.ToString();``                ``}``            ``}``        ``}` `        ``// Sort and print the output string``        ``Console.WriteLine(output);``    ``}` `    ``// Driver code``    ``static` `void` `Main(``string``[] args)``    ``{``        ``int` `n = 3225;``        ``PrintNumber(n);``    ``}``}`

## Javascript

 `// Function to print such a number``function` `printNumber(n) {``    ``// Count the frequency of each digit``    ``let freq = ``new` `Map();``    ``while` `(n > 0) {``        ``freq.set(n % 10, (freq.get(n % 10) || 0) + 1);``        ``n = Math.floor(n / 10);``    ``}` `    ``// Construct the output string``    ``let output = ``""``;``    ``for` `(let i = 1; i <= 9; i++) {``        ``if` `(freq.get(i) > 0) {``            ``for` `(let j = 0; j < freq.get(i) * i; j++) {``                ``output += i.toString();``            ``}``        ``}``    ``}` `    ``// Sort and print the output string``    ``output = output.split(``''``).sort().join(``''``);``    ``console.log(output);``}` `// Driver code``let n = 3225;``printNumber(n);`

Output

`222233355555`

Time Complexity: O(N LOG N)

Auxiliary Space: O(N)

Approach:
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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## Javascript

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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.

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