从具有相同位数的给定大数中找出可能的最小数

给定一个长度为N的数字K ，任务是通过任意次数交换数字，找到可以由 N 个数字中的 K 个组成的最小可能数字。
例子：

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 
 

方法：想法是使用哈希。为了实现散列，创建了一个大小为 10 的数组arr[] 。迭代给定的数字，并将每个数字的出现次数存储在相应索引处的散列中。然后迭代哈希数组并根据其频率打印第i 个数字。输出将是所需的最小 N 位数。
下面是上述方法的实现：

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[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
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

 
// Javscript 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));
 

输出：

112233333344457

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