给定一个由N 个正整数和一个整数K组成的数组arr[] ，任务是找到需要被其他数组元素替换的数组元素的最少数量，使得该数组最多包含K 个不同的元素。
Input: arr[] = { 1, 1, 2, 2, 5 }, K = 2
Output: 1
Explanation:
Replacing arr[4] with arr[0] modifies arr[] to { 1, 1, 2, 2, 1 }
Distinct array elements of the array arr[] are { 1, 2 }
Therefore, the required output is 1.
Input: arr[] = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 }, K = 3
Output: 3
Explanation:
Replacing arr[0] with arr[1] modifies arr[] to { 1, 1, 3, 2, 4, 1, 1, 2, 3, 4 }
Replacing arr[2] with arr[0] modifies arr[] to { 1, 1, 1, 2, 4, 1, 1, 2, 3, 4 }
Replacing arr[8] with arr[0] modifies arr[] to { 1, 1, 1, 2, 4, 1, 1, 2, 1, 4 }
Distinct array elements of the array arr[] are { 1, 2, 4 }
Therefore, the required output is 3.
方法：该问题可以使用贪心技术解决。这个想法是用更高频率的阵列元件替换较小频率的阵列元件。请按照以下步骤解决问题：
- 初始化一个映射，比如mp ，以存储每个不同数组元素的频率。
- 遍历数组并将数组中每个不同元素的频率存储到 Map 中。
- 使用地图的键值作为i遍历地图，并将mp[i]的值插入到一个数组中，比如Freq[] 。
- 按降序对数组Freq[]进行排序。
- 初始化一个变量，比如cntMin ，以存储需要替换的数组元素的最小计数，以便数组中不同元素的计数最多为K 。
- 初始化一个变量，比如len ，以存储数组Freq[]的大小。
- 使用变量i迭代范围[K, len] 。对于每个第i个值，更新cntMin += Freq[i] 。
- 最后，打印cntMin的值。
下面是上述方法的实现：
C++
// C++ program to implement
// the above approach
#include
using namespace std;
// Function to find minimum count of array
// elements required to be replaced such that
// count of distinct elements is at most K
int min_elements(int arr[], int N, int K)
{
// Store the frequency of each
// distinct element of the array
map mp;
// Traverse the array
for (int i = 0; i < N; i++) {
// Update frequency
// of arr[i]
mp[arr[i]]++;
}
// Store frequency of each distinct
// element of the array
vector Freq;
// Traverse the map
for (auto it : mp) {
// Stores key of the map
int i = it.first;
// Insert mp[i] into Freq[]
Freq.push_back(mp[i]);
}
// Sort Freq[] in descending order
sort(Freq.rbegin(), Freq.rend());
// Stores size of Freq[]
int len = Freq.size();
// If len is less than
// or equal to K
if (len <= K) {
return 0;
}
// Stores minimum count of array elements
// required to be replaced such that
// count of distinct elements is at most K
int cntMin = 0;
// Iterate over the range [K, len]
for (int i = K; i < len; i++) {
// Update cntMin
cntMin += Freq[i];
}
return cntMin;
}
// Driver Code
int main()
{
int arr[] = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 };
int N = sizeof(arr) / sizeof(arr[0]);
int K = 3;
cout << min_elements(arr, N, K);
return 0;
} Java
// Java program to implement
// the above approach
import java.util.*;
class GFG
{
// Function to find minimum count of array
// elements required to be replaced such that
// count of distinct elements is at most K
static int min_elements(int arr[], int N, int K)
{
// Store the frequency of each
// distinct element of the array
HashMap mp = new HashMap();
// Traverse the array
for (int i = 0; i < N; i++)
{
// Update frequency
// of arr[i]
if(mp.containsKey(arr[i]))
{
mp.put(arr[i], mp.get(arr[i])+1);
}
else
{
mp.put(arr[i], 1);
}
}
// Store frequency of each distinct
// element of the array
Vector Freq = new Vector();
// Traverse the map
for (Map.Entry it : mp.entrySet())
{
// Stores key of the map
int i = it.getKey();
// Insert mp[i] into Freq[]
Freq.add(mp.get(i));
}
// Sort Freq[] in descending order
Collections.sort(Freq,Collections.reverseOrder());
// Stores size of Freq[]
int len = Freq.size();
// If len is less than
// or equal to K
if (len <= K)
{
return 0;
}
// Stores minimum count of array elements
// required to be replaced such that
// count of distinct elements is at most K
int cntMin = 0;
// Iterate over the range [K, len]
for (int i = K; i < len; i++)
{
// Update cntMin
cntMin += Freq.get(i);
}
return cntMin;
}
// Driver Code
public static void main(String[] args)
{
int arr[] = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 };
int N = arr.length;
int K = 3;
System.out.print(min_elements(arr, N, K));
}
}
// This code is contributed by shikhasingrajput Python3
# Python3 program to implement
# the above approach
# Function to find minimum count of array
# elements required to be replaced such that
# count of distinct elements is at most K
def min_elements(arr, N, K) :
# Store the frequency of each
# distinct element of the array
mp = {}
# Traverse the array
for i in range(N) :
# Update frequency
# of arr[i]
if arr[i] in mp :
mp[arr[i]] += 1
else :
mp[arr[i]] = 1
# Store frequency of each distinct
# element of the array
Freq = []
# Traverse the map
for it in mp :
# Stores key of the map
i = it
# Insert mp[i] into Freq[]
Freq.append(mp[i])
# Sort Freq[] in descending order
Freq.sort()
Freq.reverse()
# Stores size of Freq[]
Len = len(Freq)
# If len is less than
# or equal to K
if (Len <= K) :
return 0
# Stores minimum count of array elements
# required to be replaced such that
# count of distinct elements is at most K
cntMin = 0
# Iterate over the range [K, len]
for i in range(K, Len) :
# Update cntMin
cntMin += Freq[i]
return cntMin
# Driver code
arr = [ 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 ]
N = len(arr)
K = 3;
print(min_elements(arr, N, K))
# This code is contributed by divyesh072019.C#
// C# program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG{
// Function to find minimum count of array
// elements required to be replaced such that
// count of distinct elements is at most K
static int min_elements(int []arr, int N, int K)
{
// Store the frequency of each
// distinct element of the array
Dictionary mp = new Dictionary();
// Traverse the array
for(int i = 0; i < N; i++)
{
// Update frequency
// of arr[i]
if (mp.ContainsKey(arr[i]))
{
mp[arr[i]] = mp[arr[i]] + 1;
}
else
{
mp.Add(arr[i], 1);
}
}
// Store frequency of each distinct
// element of the array
List Freq = new List();
// Traverse the map
foreach (KeyValuePair it in mp)
{
// Stores key of the map
int i = it.Key;
// Insert mp[i] into Freq[]
Freq.Add(mp[i]);
}
// Sort Freq[] in descending order
Freq.Sort();
Freq.Reverse();
// Stores size of Freq[]
int len = Freq.Count;
// If len is less than
// or equal to K
if (len <= K)
{
return 0;
}
// Stores minimum count of array elements
// required to be replaced such that
// count of distinct elements is at most K
int cntMin = 0;
// Iterate over the range [K, len]
for(int i = K; i < len; i++)
{
// Update cntMin
cntMin += Freq[i];
}
return cntMin;
}
// Driver Code
public static void Main(String[] args)
{
int []arr = { 5, 1, 3, 2, 4, 1, 1, 2, 3, 4 };
int N = arr.Length;
int K = 3;
Console.Write(min_elements(arr, N, K));
}
}
// This code is contributed by gauravrajput1 输出：
3时间复杂度： O(N * log(N))
辅助空间： O(N)
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