将数组拆分为K个长度的子集，以最小化每个子集的第二个最小元素的总和

给定大小为N的数组arr []和整数K ( N％K = 0 )，任务是将数组拆分为大小为K的子数组，以使每个子数组的^第二个最小元素的总和为最小。

例子：

Input: arr[] = {11, 20, 5, 7, 8, 14, 2, 17, 16, 10}, K = 5
Output: 13
Explanation: Splitting array into subsets {11, 5, 14, 2, 10} and {20, 7, 8, 17, 16} generates minimum sum of second smallest elements( = 5 + 8 = 13).

Input: arr[] = {13, 19, 20, 5, 17, 11, 10, 8, 23, 14}, K = 5
Output: 19
Explanation: Splitting array into subsets {10, 19, 20, 11, 23} and {17, 5, 8, 13, 14} generates minimum sum of second smallest elements(= 11 + 8 = 19).

方法：可以通过排序技术解决问题。请按照以下步骤解决此问题：

• 以升序对给定数组进行排序。
• 由于所有子集的长度均为K组，因此请从1开始选择前K个奇数索引元素，然后计算它们的总和。
• 打印获得的总和。

下面是上述方法的实现：

C++

// C++ implementation for above approach
  
#include 
using namespace std;
  
// Function to find the minimum sum of
// 2nd smallest elements of each subset
int findMinimum(int arr[], int N, int K)
{
    // Sort the array
    sort(arr, arr + N);
  
    // Stores minimum sum of second
    // elements of each subset
    int ans = 0;
  
    // Traverse first K 2nd smallest elements
    for (int i = 1; i < 2 * (N / K); i += 2) {
  
        // Update their sum
        ans += arr[i];
    }
  
    // Print the sum
    cout << ans;
}
  
// Driver Code
int main()
{
    // Given Array
    int arr[] = { 11, 20, 5, 7, 8,
                  14, 2, 17, 16, 10 };
  
    // Given size of the array
    int N = sizeof(arr)
            / sizeof(arr[0]);
  
    // Given subset lengths
    int K = 5;
  
    findMinimum(arr, N, K);
  
    return 0;
}

Java

// Java implementation for the above approach
  
import java.io.*;
import java.util.*;
  
class GFG {
  
    // Function to find the minimum sum of
    // 2nd smallest elements of each subset
    public static void findMinimum(
        int arr[], int N, int K)
    {
        // Sort the array
        Arrays.sort(arr);
  
        // Stores minimum sum of second
        // elements of each subset
        int ans = 0;
  
        // Traverse first K 2nd smallest elements
        for (int i = 1; i < 2 * (N / K); i += 2) {
  
            // Update their sum
            ans += arr[i];
        }
  
        // Print the sum
        System.out.println(ans);
    }
  
    // Driver Code
    public static void main(String[] args)
    {
        // Given Array
        int[] arr = { 11, 20, 5, 7, 8,
                      14, 2, 17, 16, 10 };
  
        // Given length of array
        int N = arr.length;
  
        // Given subset lengths
        int K = 5;
  
        findMinimum(arr, N, K);
    }
}

Python3

# Python3 implementation for above approach
  
# Function to find the minimum sum of
# 2nd smallest elements of each subset
def findMinimum(arr, N, K):
    
    # Sort the array
    arr = sorted(arr)
  
    # Stores minimum sum of second
    # elements of each subset
    ans = 0
  
    # Traverse first K 2nd smallest elements
    for i in range(1, 2 * (N//K), 2):
  
        # Update their sum
        ans += arr[i]
  
    # Prthe sum
    print (ans)
  
# Driver Code
if __name__ == '__main__':
    
    # Given Array
    arr = [11, 20, 5, 7, 8, 14, 2, 17, 16, 10]
  
    # Given size of the array
    N = len(arr)
  
    # Given subset lengths
    K = 5
    findMinimum(arr, N, K)
  
# This code is contributed by mohit kumar 29

C#

// C# implementation for above approach 
using System;
   
class GFG{
   
// Function to find the minimum sum of
// 2nd smallest elements of each subset
public static void findMinimum(int[] arr, int N,
                               int K)
{
      
    // Sort the array
    Array.Sort(arr);
  
    // Stores minimum sum of second
    // elements of each subset
    int ans = 0;
  
    // Traverse first K 2nd smallest elements
    for(int i = 1; i < 2 * (N / K); i += 2) 
    {
          
        // Update their sum
        ans += arr[i];
    }
  
    // Print the sum
    Console.WriteLine(ans);
}
  
// Driver Code
public static void Main()
{
      
    // Given Array
    int[] arr = { 11, 20, 5, 7, 8,
                  14, 2, 17, 16, 10 };
  
    // Given length of array
    int N = arr.Length;
  
    // Given subset lengths
    int K = 5;
  
    findMinimum(arr, N, K);
}
}
  
// This code is contributed by susmitakundugoaldanga

输出：

13

时间复杂度： O(NlogN)
空间复杂度： O(N)