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

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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

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:

C++

// C++ implementation of the above approach
 
#include <iostream>
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[10] = { 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
import java.util.*;
import java.io.*;
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[10];
 
    // 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

Javascript

<script>
 
// Javascript implementation of the above approach
 
// Function for finding the smallest
// possible number after swapping
// the digits any number of times
function smallestPoss(s, n)
{
    // Variable to store the final answer
    var ans = "";
 
    // Array to store the count of
    // occurrence of each digit
    var arr = Array(10).fill(0);
 
    // Loop to calculate the number
    // of occurrences of every digit
    for (var i = 0; i < n; i++) {
        arr[s[i].charCodeAt(0) - 48]++;
    }
 
    // Loop to get smallest number
    for (var i = 0; i < 10; i++) {
        for (var j = 0; j < arr[i]; j++)
            ans = ans + i.toString();
    }
 
    // Returning the answer
    return ans;
}
 
// Driver code
var N = 15;
var K = "325343273113434";
document.write( smallestPoss(K, N));
 
</script>

Output:

112233333344457

Time Complexity: O(N)

Auxiliary Space: O(N + 10)

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