找出一个数组中包含 k 个不同元素所需的最小变化
原文:https://www . geeksforgeeks . org/find-最小化-更改-要求在数组中包含 k 个不同的元素/
给定一个大小为 N 的数组和一个数量为 K 的数组。任务是找到数组中要替换的任意数量的最小元素，使数组由 K 个不同的元素组成。 注意:数组可能由重复元素组成。 举例:
输入: arr[]={1，2，2，8}，k = 1 输出: 2 需要更改的元素为 1，8 输入: arr[]={1，2，7，8，2，3，2，3}，k = 2 输出: 3 需要更改的元素为 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。所以定义一个数组freq【】存储数组 arr 中每个数字的频率，然后按降序排序 freq 。所以，首先 freq 阵列的 k 元素不需要更换。 以下是上述方法的实施:
C++
// CPP program to minimum changes required
// in an array for k distinct elements.
#include <bits/stdc++.h>
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<int>());
// 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 语言(一种计算机语言，尤用于创建网站)
// 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
Python 3
# 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
java 描述语言
<script>
// Javascript program to minimum changes required
// in an array for k distinct elements.
var MAX = 100005;
// Function to minimum changes required
// in an array for k distinct elements.
function Min_Replace(arr, n, k)
{
arr.sort((a,b)=>a-b)
// Store the frequency of each element
var freq = Array(MAX).fill(0);
var p = 0;
freq[p] = 1;
// Store the frequency of elements
for (var i = 1; i < n; i++) {
if (arr[i] == arr[i - 1])
++freq[p];
else
++freq[++p];
}
// Sort frequencies in descending order
freq.sort((a,b)=>b-a);
// To store the required answer
var ans = 0;
for (var i = k; i <= p; i++)
ans += freq[i];
// Return the required answer
return ans;
}
// Driver code
var arr = [1, 2, 7, 8, 2, 3, 2, 3];
var n = arr.length;
var k = 2;
document.write( Min_Replace(arr, n, k));
</script>
输出:
3
时间复杂度: O(NlogN)
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