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
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++
#include <iostream>
using namespace std;
string smallestPoss(string s, int n)
{
string ans = "" ;
int arr[10] = { 0 };
for ( int i = 0; i < n; i++) {
arr[s[i] - 48]++;
}
for ( int i = 0; i < 10; i++) {
for ( int j = 0; j < arr[i]; j++)
ans = ans + to_string(i);
}
return ans;
}
int main()
{
int N = 15;
string K = "325343273113434" ;
cout << smallestPoss(K, N);
return 0;
}
|
Java
import java.util.*;
import java.io.*;
class GFG
{
static String smallestPoss(String s, int n)
{
String ans = "" ;
int arr[] = new int [ 10 ];
for ( int i = 0 ; i < n; i++)
{
arr[s.charAt(i) - 48 ]++;
}
for ( int i = 0 ; i < 10 ; i++)
{
for ( int j = 0 ; j < arr[i]; j++)
ans = ans + String.valueOf(i);
}
return ans;
}
public static void main(String[] args)
{
int N = 15 ;
String K = "325343273113434" ;
System.out.print(smallestPoss(K, N));
}
}
|
Python3
def smallestPoss(s, n):
ans = "";
arr = [ 0 ] * 10 ;
for i in range (n):
arr[ ord (s[i]) - 48 ] + = 1 ;
for i in range ( 10 ):
for j in range (arr[i]):
ans = ans + str (i);
return ans;
if __name__ = = '__main__' :
N = 15 ;
K = "325343273113434" ;
print (smallestPoss(K, N));
|
C#
using System;
class GFG
{
static String smallestPoss(String s, int n)
{
String ans = "" ;
int []arr = new int [10];
for ( int i = 0; i < n; i++)
{
arr[s[i] - 48]++;
}
for ( int i = 0; i < 10; i++)
{
for ( int j = 0; j < arr[i]; j++)
ans = ans + String.Join( "" ,i);
}
return ans;
}
public static void Main(String[] args)
{
int N = 15;
String K = "325343273113434" ;
Console.Write(smallestPoss(K, N));
}
}
|
Javascript
<script>
function smallestPoss(s, n)
{
var ans = "" ;
var arr = Array(10).fill(0);
for ( var i = 0; i < n; i++) {
arr[s[i].charCodeAt(0) - 48]++;
}
for ( var i = 0; i < 10; i++) {
for ( var j = 0; j < arr[i]; j++)
ans = ans + i.toString();
}
return ans;
}
var N = 15;
var K = "325343273113434" ;
document.write( smallestPoss(K, N));
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N + 10)
Last Updated :
20 Dec, 2022
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...