# Find smallest possible Number from a given large Number with same count of digits

Given a number K of length N, the task is to find the smallest possible number that can be formed from K of N digits by swapping the digits any number of times.

Examples:

Input: N = 15, K = 325343273113434
Output: 112233333344457
Explanation:
The smallest number possible after swapping the digits of the given number is 112233333344457

Input: N = 7, K = 3416781
Output: 1134678

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

Approach: The idea is to use Hashing. To implement the hash, an array arr[] of size 10 is created. The given number is iterated and the count of occurrence of every digit is stored in the hash at the corresponding index. Then iterate the hash array and print the ith digit according to its frequency. The output will be the smallest required number of N digits.

Below is the implementation of the above approach:

## CPP

 `// C++ implementation of the above approach ` ` `  `#include ` `using` `namespace` `std; ` ` `  `// Function for finding the smallest ` `// possible number after swapping ` `// the digits any number of times ` `string smallestPoss(string s, ``int` `n) ` `{ ` `    ``// Variable to store the final answer ` `    ``string ans = ``""``; ` ` `  `    ``// Array to store the count of ` `    ``// occurrence of each digit ` `    ``int` `arr = { 0 }; ` ` `  `    ``// Loop to calculate the number ` `    ``// of occurrences of every digit ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``arr[s[i] - 48]++; ` `    ``} ` ` `  `    ``// Loop to get smallest number ` `    ``for` `(``int` `i = 0; i < 10; i++) { ` `        ``for` `(``int` `j = 0; j < arr[i]; j++) ` `            ``ans = ans + to_string(i); ` `    ``} ` ` `  `    ``// Returning the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` `    ``int` `N = 15; ` `    ``string K = ``"325343273113434"``; ` ` `  `    ``cout << smallestPoss(K, N); ` ` `  `    ``return` `0; ` `} `

## Java

 `// Java implementation of the above approach ` `class` `GFG ` `{ ` ` `  `// Function for finding the smallest ` `// possible number after swapping ` `// the digits any number of times ` `static` `String smallestPoss(String s, ``int` `n) ` `{ ` `    ``// Variable to store the final answer ` `    ``String ans = ``""``; ` ` `  `    ``// Array to store the count of ` `    ``// occurrence of each digit ` `    ``int` `arr[] = ``new` `int``[``10``]; ` ` `  `    ``// Loop to calculate the number ` `    ``// of occurrences of every digit ` `    ``for` `(``int` `i = ``0``; i < n; i++) ` `    ``{ ` `        ``arr[s.charAt(i) - ``48``]++; ` `    ``} ` ` `  `    ``// Loop to get smallest number ` `    ``for` `(``int` `i = ``0``; i < ``10``; i++) ` `    ``{ ` `        ``for` `(``int` `j = ``0``; j < arr[i]; j++) ` `            ``ans = ans + String.valueOf(i); ` `    ``} ` ` `  `    ``// Returning the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `main(String[] args) ` `{ ` `    ``int` `N = ``15``; ` `    ``String K = ``"325343273113434"``; ` ` `  `    ``System.out.print(smallestPoss(K, N)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

## Python3

 `# Python3 implementation of the above approach ` ` `  `# Function for finding the smallest ` `# possible number after swapping ` `# the digits any number of times ` `def` `smallestPoss(s, n): ` `     `  `    ``# Variable to store the final answer ` `    ``ans ``=` `""; ` ` `  `    ``# Array to store the count of ` `    ``# occurrence of each digit ` `    ``arr ``=` `[``0``]``*``10``; ` ` `  `    ``# Loop to calculate the number ` `    ``# of occurrences of every digit ` `    ``for` `i ``in` `range``(n): ` `        ``arr[``ord``(s[i]) ``-` `48``] ``+``=` `1``; ` `     `  `    ``# Loop to get smallest number ` `    ``for` `i ``in` `range``(``10``): ` `        ``for` `j ``in` `range``(arr[i]): ` `            ``ans ``=` `ans ``+` `str``(i); ` `     `  `    ``# Returning the answer ` `    ``return` `ans; ` ` `  `# Driver code ` `if` `__name__ ``=``=` `'__main__'``: ` `    ``N ``=` `15``; ` `    ``K ``=` `"325343273113434"``; ` ` `  `    ``print``(smallestPoss(K, N)); ` ` `  `# This code is contributed by 29AjayKumar `

## C#

 `// C# implementation of the above approach ` `using` `System; ` ` `  `class` `GFG ` `{ ` ` `  `// Function for finding the smallest ` `// possible number after swapping ` `// the digits any number of times ` `static` `String smallestPoss(String s, ``int` `n) ` `{ ` `    ``// Variable to store the readonly answer ` `    ``String ans = ``""``; ` ` `  `    ``// Array to store the count of ` `    ``// occurrence of each digit ` `    ``int` `[]arr = ``new` `int``; ` ` `  `    ``// Loop to calculate the number ` `    ``// of occurrences of every digit ` `    ``for` `(``int` `i = 0; i < n; i++) ` `    ``{ ` `        ``arr[s[i] - 48]++; ` `    ``} ` ` `  `    ``// Loop to get smallest number ` `    ``for` `(``int` `i = 0; i < 10; i++) ` `    ``{ ` `        ``for` `(``int` `j = 0; j < arr[i]; j++) ` `            ``ans = ans + String.Join(``""``,i); ` `    ``} ` ` `  `    ``// Returning the answer ` `    ``return` `ans; ` `} ` ` `  `// Driver code ` `public` `static` `void` `Main(String[] args) ` `{ ` `    ``int` `N = 15; ` `    ``String K = ``"325343273113434"``; ` ` `  `    ``Console.Write(smallestPoss(K, N)); ` `} ` `} ` ` `  `// This code is contributed by PrinciRaj1992 `

Output:

```112233333344457
```

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : princiraj1992, 29AjayKumar