从给定的大数字中找出具有相同位数的最小可能数字

给定长度为N的数字K ，任务是通过将数字交换任意次数来找到可以由N个数字中的K个组成的最小的可能数字。

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。为了实现哈希，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哈希表中的相应索引处。然后迭代哈希数组，并根据其频率打印第i个数字。输出将是所需的最少N位数字。

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

输出：

112233333344457