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 112233333344457Input: 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 timesstring 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 codeint main(){ int N = 15; string K = "325343273113434"; cout << smallestPoss(K, N); return 0;} |
Java
// Java implementation of the above approachimport java.util.*;import java.io.*;class GFG{// Function for finding the smallest// possible number after swapping// the digits any number of timesstatic 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 codepublic 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 timesdef 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 codeif __name__ == '__main__': N = 15; K = "325343273113434"; print(smallestPoss(K, N));# This code is contributed by 29AjayKumar |
C#
// C# implementation of the above approachusing System;class GFG{// Function for finding the smallest// possible number after swapping// the digits any number of timesstatic 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 codepublic 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 timesfunction 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 codevar 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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