最大化给定字符串中 K 个选定字符的频率总和

给定一个正整数K和一个由N 个字符组成的字符串S ，任务是最大化从给定字符串中恰好选择 K 个字符的频率之和。

例子：

Input: K – 3, S = ”GEEKSFORGEEKS”
Output: 12
Explanation:
Choose the character E from the given string S, the sum of frequency of 3 E’s is 3*4 = 12, which is the maximum.

Input: K = 10, S = ”GEEKSFORGEEKS”
Output: 28

方法：通过选择具有较高频率的K 个字符，可以使用贪心方法解决给定的问题。请按照以下步骤解决问题：

• 初始化一个变量，比如sum为0 ，它存储从字符串S 中选择的恰好 K 个字符的频率的结果总和。
• 找到字符串S中每个字符的频率并将其存储在一个辅助数组中，比如freq[] 。
• 按降序对数组freq[]进行排序。
• 遍历数组freq[]并执行以下操作：
  • 如果当前频率小于K ，则将freq[i]*freq[i] 添加到变量sum 中，因为它最大化结果总和，并在选择freq[i]字符从K递减freq[i]的值。
  • 否则，将K*freq[i]的值添加到变量sum并在选择了K 个字符跳出循环。
• 完成上述步骤后，打印sum的值作为结果。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
 
// Function to find the maximum sum of
// frequencies of the exactly K chosen
// characters from the string S
int maximumSum(string S, int N, int K)
{
    // Stores the resultant maximum sum
    int sum = 0;
 
    // Stores the frequency of array
    // elements
    int freq[256] = { 0 };
 
    // Find the frequency of character
    for (int i = 0; i < N; i++) {
        freq[int(S[i])]++;
    }
 
    // Sort the frequency array in the
    // descending order
    sort(freq, freq + 256, greater());
 
    // Iterate to chose K elements greedily
    for (int i = 0; i < 256; i++) {
 
        // If the freq[i] cards are
        // chosen
        if (K > freq[i]) {
            sum += freq[i] * freq[i];
            K -= freq[i];
        }
 
        // K cards have been picked
        else {
            sum += freq[i] * K;
            break;
        }
    }
 
    // Return the resultant sum
    return sum;
}
 
// Driver Code
int main()
{
    string S = "GEEKSFORGEEKS";
    int K = 10;
    int N = S.length();
    cout << maximumSum(S, N, K);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
class GFG
{
 
// Function to find the maximum sum of
// frequencies of the exactly K chosen
// characters from the String S
static int maximumSum(String S, int N, int K)
{
     
    // Stores the resultant maximum sum
    int sum = 0;
 
    // Stores the frequency of array
    // elements
    Integer []freq = new Integer[256];
    Arrays.fill(freq, 0);
 
    // Find the frequency of character
    for(int i = 0; i < N; i++)
    {
        freq[(int)S.charAt(i)] += 1;
    }
 
    // Sort the frequency array in the
    // descending order
     Arrays.sort(freq, Collections.reverseOrder());
 
    // Iterate to chose K elements greedily
    for(int i = 0; i < 256; i++)
    {
         
        // If the freq[i] cards are
        // chosen
        if (K > freq[i])
        {
            sum += freq[i] * freq[i];
            K -= freq[i];
        }
 
        // K cards have been picked
        else
        {
            sum += freq[i] * K;
            break;
        }
    }
 
    // Return the resultant sum
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    String S = "GEEKSFORGEEKS";
    int K = 10;
    int N = S.length();
     
    System.out.print(maximumSum(S, N, K));
}
}
 
// This code is contributed by ipg2016107

Python3

# Python3 program for the above approach
 
# Function to find the maximum sum of
# frequencies of the exactly K chosen
# characters from the string S
def maximumSum(S, N, K):
     
    # Stores the resultant maximum sum
    sum = 0
 
    # Stores the frequency of array
    # elements
    freq = [0] * 256
     
    # Find the frequency of character
    for i in range(N):
        freq[ord(S[i])] += 1
 
    # Sort the frequency array in the
    # descending order
    freq = sorted(freq)[::-1]
 
    # Iterate to chose K elements greedily
    for i in range(256):
         
        # If the freq[i] cards are
        # chosen
        if (K > freq[i]):
            sum += freq[i] * freq[i]
            K -= freq[i]
             
        # K cards have been picked
        else:
            sum += freq[i] * K
            break
 
    # Return the resultant sum
    return sum
 
# Driver Code
if __name__ == '__main__':
     
    S = "GEEKSFORGEEKS"
    K = 10
    N = len(S)
     
    print(maximumSum(S, N, K))
 
# This code is contributed by mohit kumar 29

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
  
// Function to find the maximum sum of
// frequencies of the exactly K chosen
// characters from the string S
static int maximumSum(string S, int N, int K)
{
     
    // Stores the resultant maximum sum
    int sum = 0;
 
    // Stores the frequency of array
    // elements
    int []freq = new int[256];
    Array.Clear(freq, 0, 256);
 
    // Find the frequency of character
    for(int i = 0; i < N; i++)
    {
        freq[(int)S[i]]++;
    }
 
    // Sort the frequency array in the
    // descending order
    Array.Sort(freq);
    Array.Reverse(freq);
 
    // Iterate to chose K elements greedily
    for(int i = 0; i < 256; i++)
    {
         
        // If the freq[i] cards are
        // chosen
        if (K > freq[i])
        {
            sum += freq[i] * freq[i];
            K -= freq[i];
        }
 
        // K cards have been picked
        else
        {
            sum += freq[i] * K;
            break;
        }
    }
 
    // Return the resultant sum
    return sum;
}
 
// Driver Code
public static void Main()
{
    string S = "GEEKSFORGEEKS";
    int K = 10;
    int N = S.Length;
     
    Console.Write(maximumSum(S, N, K));
}
}
 
// This code is contributed by ipg2016107

Javascript

 
// JavaScript program for the above approach
 
// Function to find the maximum sum of
// frequencies of the exactly K chosen
// characters from the string S
function maximumSum(S, N, K)
{
     
    // Stores the resultant maximum sum
    let sum = 0;
 
    // Stores the frequency of array
    // elements
    let freq = Array.from({length: 256}, (_, i) => 0);
 
    // Find the frequency of character
    for(let i = 0; i < N; i++)
    {
        freq[S[i].charCodeAt()]++;
    }
 
    // Sort the frequency array in the
    // descending order
    freq.sort((a, b) => b - a);
 
    // Iterate to chose K elements greedily
    for(let i = 0; i < 256; i++)
    {
         
        // If the freq[i] cards are
        // chosen
        if (K > freq[i])
        {
            sum += freq[i] * freq[i];
            K -= freq[i];
        }
 
        // K cards have been picked
        else
        {
            sum += freq[i] * K;
            break;
        }
    }
 
    // Return the resultant sum
    return sum;
}       
 
// Driver Code
 
    let S = "GEEKSFORGEEKS";
    let K = 10;
    let N = S.length;
     
    document.write(maximumSum(S.split(''), N, K))
 

输出：

28

时间复杂度： O(N)
辅助空间： O(26)

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