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

• Last Updated : 20 Dec, 2022

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 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



Output:

112233333344457

Time Complexity: O(N)

Auxiliary Space: O(N + 10)

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