Split array into K-length subsets to minimize sum of second smallest element of each subset

Last Updated : 18 Jan, 2022

Given an array arr[] of size N and an integer K (N % K = 0), the task is to split array into subarrays of size K such that the sum of 2^nd smallest elements of each subarray is the minimum possible.

Examples:

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).

Approach: The problem can be solved by sorting technique. Follow the steps below to solve this problem:

Sort the given array in ascending order.
Since all subsets are of length Kgroups, choose first K odd-indexed elements starting from 1 and calculate their sum.
Print the obtained sum.

Below is the implementation of the above approach:

C++

// C++ implementation for above approach

#include <bits/stdc++.h>
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]

    # Print the 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# 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

JavaScript

<script>

// Javascript implementation for
// the above approach

    // Function to find the minimum sum of
    // 2nd smallest elements of each subset
    function findMinimum(arr , N , K) 
    {
        // Sort the array
        arr.sort((a,b)=>a-b);

        // Stores minimum sum of second
        // elements of each subset
        var ans = 0;

        // Traverse first K 2nd smallest elements
        for (i = 1; i < 2 * (parseInt(N / K)); i += 2) 
        {

            // Update their sum
            ans += arr[i];
        }

        // Print the sum
        document.write(ans);
    }

    // Driver Code
    
        // Given Array
        var arr = [ 11, 20, 5, 7, 8, 14, 2, 17, 16, 10 ];

        // Given length of array
        var N = arr.length;

        // Given subset lengths
        var K = 5;

        findMinimum(arr, N, K);

// This code is contributed by todaysgaurav

</script>

Output:

Time Complexity: O(NlogN)
Space Complexity: O(N)

Smallest subset of maximum sum possible by splitting array into two subsets

RohitOberoi

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Split array into K-length subsets to minimize sum of second smallest element of each subset

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