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📜  将数组拆分为 K 个子集以最大化其第二大元素的总和

📅  最后修改于: 2021-09-06 05:55:08             🧑  作者: Mango

给定一个由N 个整数和一个整数K组成的数组arr[] ,任务是将数组拆分为K个子集( N % K = 0 ),使得所有子集的第二大元素的总和最大化。

例子:

方法:想法是对数组进行排序,并在反向遍历数组时,从数组中的第二大元素开始,恰好K次,将遇到的每个第二个元素相加。请按照以下步骤解决问题:

  • 按升序对数组进行排序。
  • 反向遍历数组arr[]
  • 初始化i = N – 1
  • 迭代K次并将arr[i – 1]添加到总和并将i减少2
  • 最后,打印得到的总和。

下面是上述方法的实现:

C++
// C++ program to implement
// the above approach
 
#include 
using namespace std;
 
// Function to split array into
// K subsets having maximum
// sum of their second maximum elements
void splitArray(int arr[], int n, int K)
{
    // Sort the array
    sort(arr, arr + n);
 
    int i = n - 1;
 
    // Stores the maximum possible
    // sum of second maximums
    int result = 0;
 
    while (K--) {
 
        // Add second maximum
        // of current subset
        result += arr[i - 1];
 
        // Proceed to the second
        // maximum of next subset
        i -= 2;
    }
 
    // Print the maximum
    // sum obtained
    cout << result;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 3, 1, 5, 1, 3 };
 
    // Size of array
    int N = sizeof(arr)
            / sizeof(arr[0]);
 
    int K = 2;
 
    // Function Call
    splitArray(arr, N, K);
 
    return 0;
}


Java
// Java program for the above approach
import java.io.*;
import java.util.Arrays;
   
class GFG
{
   
// Function to split array into
// K subsets having maximum
// sum of their second maximum elements
static void splitArray(int arr[], int n, int K)
{
    // Sort the array
    Arrays.sort(arr);
  
    int i = n - 1;
  
    // Stores the maximum possible
    // sum of second maximums
    int result = 0;
  
    while (K-- != 0)
    {
  
        // Add second maximum
        // of current subset
        result += arr[i - 1];
  
        // Proceed to the second
        // maximum of next subset
        i -= 2;
    }
  
    // Print the maximum
    // sum obtained
    System.out.print(result);
}
   
// Drive Code
public static void main(String[] args)
{
    // Given array arr[]
    int[] arr = { 1, 3, 1, 5, 1, 3 };
  
    // Size of array
    int N = arr.length;
  
    int K = 2;
  
    // Function Call
    splitArray(arr, N, K);
}
}
 
// This code is contributed by sanjoy_62.


Python3
# Python3 program to implement
# the above approach
 
# Function to split array into K
# subsets having maximum sum of
# their second maximum elements
def splitArray(arr, n, K):
     
    # Sort the array
    arr.sort()
 
    i = n - 1
 
    # Stores the maximum possible
    # sum of second maximums
    result = 0
 
    while (K > 0):
 
        # Add second maximum
        # of current subset
        result += arr[i - 1]
 
        # Proceed to the second
        # maximum of next subset
        i -= 2
        K -= 1
 
    # Print the maximum
    # sum obtained
    print(result)
 
# Driver Code
if __name__ == "__main__":
 
    # Given array arr[]
    arr = [ 1, 3, 1, 5, 1, 3 ]
 
    # Size of array
    N = len(arr)
 
    K = 2
 
    # Function Call
    splitArray(arr, N, K)
 
# This code is contributed by chitranayal


C#
// C# program for the above approach
using System;
class GFG
{
   
// Function to split array into
// K subsets having maximum
// sum of their second maximum elements
static void splitArray(int []arr, int n, int K)
{
    // Sort the array
    Array.Sort(arr);
  
    int i = n - 1;
  
    // Stores the maximum possible
    // sum of second maximums
    int result = 0;
  
    while (K-- != 0)
    {
  
        // Add second maximum
        // of current subset
        result += arr[i - 1];
  
        // Proceed to the second
        // maximum of next subset
        i -= 2;
    }
  
    // Print the maximum
    // sum obtained
    Console.Write(result);
}
   
// Drive Code
public static void Main(String[] args)
{
    // Given array []arr
    int[] arr = { 1, 3, 1, 5, 1, 3 };
  
    // Size of array
    int N = arr.Length;
  
    int K = 2;
  
    // Function Call
    splitArray(arr, N, K);
}
}
 
// This code is contributed by shikhasingrajput


Javascript


输出:
4

时间复杂度: O(N logN)
辅助空间: O(N)

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