查找数组中包含k个不同元素所需的最小更改

给定大小为N且数字为K的数组arr 。任务是在数组中找到要替换的最小元素数，以使数组包含K个不同的元素。

注意：数组可能包含重复元素。
例子：

Input : arr[]={1, 2, 2, 8}, k = 1
Output : 2
The elements to be changed are 1, 8

Input : arr[]={1, 2, 7, 8, 2, 3, 2, 3}, k = 2
Output : 3
The elements to be changed are 1, 7, 8

方法：由于任务是替换数组中的最少元素，因此我们不会替换数组中出现频率更高的元素。因此，只需定义一个数组freq []即可存储数组arr中每个数字的频率，然后按降序对freq进行排序。因此，不需要替换freq数组的前k个元素。

下面是上述方法的实现：

C++

// CPP program to minimum changes required 
// in an array for k distinct elements.
#include 
using namespace std;
  
#define MAX 100005
  
// Function to minimum changes required 
// in an array for k distinct elements.
int Min_Replace(int arr[], int n, int k)
{
    sort(arr, arr + n);
  
    // Store the frequency of each element
    int freq[MAX];
      
    memset(freq, 0, sizeof freq);
      
    int p = 0;
    freq[p] = 1;
      
    // Store the frequency of elements
    for (int i = 1; i < n; i++) {
        if (arr[i] == arr[i - 1])
            ++freq[p];
        else
            ++freq[++p];
    }
  
    // Sort frequencies in descending order
    sort(freq, freq + n, greater());
      
    // To store the required answer
    int ans = 0;
    for (int i = k; i <= p; i++)
        ans += freq[i];
          
    // Return the required answer
    return ans;
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 7, 8, 2, 3, 2, 3 };
      
    int n = sizeof(arr) / sizeof(arr[0]);
      
    int k = 2;
      
    cout << Min_Replace(arr, n, k);
      
    return 0;
}

Java

// C# program to minimum changes required 
// in an array for k distinct elements.
import java.util.*;
  
class GFG
{
    static int MAX = 100005;
      
    // Function to minimum changes required 
    // in an array for k distinct elements.
    static int Min_Replace(int [] arr, 
                           int n, int k)
    {
        Arrays.sort(arr);
      
        // Store the frequency of each element
        Integer [] freq = new Integer[MAX];
        Arrays.fill(freq, 0);
        int p = 0;
        freq[p] = 1;
          
        // Store the frequency of elements
        for (int i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                ++freq[p];
            else
                ++freq[++p];
        }
      
        // Sort frequencies in descending order
        Arrays.sort(freq, Collections.reverseOrder());
          
        // To store the required answer
        int ans = 0;
        for (int i = k; i <= p; i++)
            ans += freq[i];
              
        // Return the required answer
        return ans;
    }
      
    // Driver code
    public static void main (String []args)
    {
        int [] arr = { 1, 2, 7, 8, 2, 3, 2, 3 };
          
        int n = arr.length;
          
        int k = 2;
          
        System.out.println(Min_Replace(arr, n, k));
    }
}
  
// This code is contributed by PrinciRaj1992

Python3

# Python 3 program to minimum changes required 
# in an array for k distinct elements.
MAX = 100005
  
# Function to minimum changes required 
# in an array for k distinct elements.
def Min_Replace(arr, n, k):
    arr.sort(reverse = False)
  
    # Store the frequency of each element
    freq = [0 for i in range(MAX)]
      
    p = 0
    freq[p] = 1
      
    # Store the frequency of elements
    for i in range(1, n, 1):
        if (arr[i] == arr[i - 1]):
            freq[p] += 1
        else:
            p += 1
            freq[p] += 1
  
    # Sort frequencies in descending order
    freq.sort(reverse = True)
      
    # To store the required answer
    ans = 0
    for i in range(k, p + 1, 1):
        ans += freq[i]
          
    # Return the required answer
    return ans
  
# Driver code
if __name__ == '__main__':
    arr = [1, 2, 7, 8, 2, 3, 2, 3]
      
    n = len(arr)
      
    k = 2
      
    print(Min_Replace(arr, n, k))
      
# This code is contributed by
# Surendra_Gangwar

C#

// C# program to minimum changes required 
// in an array for k distinct elements.
using System;
  
class GFG
{
    static int MAX = 100005;
      
    // Function to minimum changes required 
    // in an array for k distinct elements.
    static int Min_Replace(int [] arr, 
                           int n, int k)
    {
        Array.Sort(arr);
      
        // Store the frequency of each element
        int [] freq = new int[MAX];
          
        int p = 0;
        freq[p] = 1;
          
        // Store the frequency of elements
        for (int i = 1; i < n; i++)
        {
            if (arr[i] == arr[i - 1])
                ++freq[p];
            else
                ++freq[++p];
        }
      
        // Sort frequencies in descending order
        Array.Sort(freq);
        Array.Reverse(freq);
          
        // To store the required answer
        int ans = 0;
        for (int i = k; i <= p; i++)
            ans += freq[i];
              
        // Return the required answer
        return ans;
    }
      
    // Driver code
    public static void Main ()
    {
        int [] arr = { 1, 2, 7, 8, 2, 3, 2, 3 };
          
        int n = arr.Length;
          
        int k = 2;
          
        Console.WriteLine(Min_Replace(arr, n, k));
    }
}
  
// This code is contributed by ihritik

输出：

3

时间复杂度： O(NlogN)