给定一个大小为N的数组arr和一个数字K 。任务是找到数组中要被任意数字替换的最小元素，使得数组由K 个不同的元素组成。
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小元素，所以我们不会替换数组中频率更高的元素。因此，只需定义一个数组freq[]来存储数组arr 中存在的每个数字的频率，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 PrinciRaj1992Python3
# 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_GangwarC#
// 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 ihritikJavascript
输出：
3时间复杂度： O(NlogN)
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