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# Split array into K subsets to maximize sum of their second largest elements

• Difficulty Level : Medium
• Last Updated : 21 Apr, 2021

Given an array arr[] consisting of N integers and an integer K, the task is to split the array into K subsets (N % K = 0) such that the sum of second largest elements of all subsets is maximized.

Examples:

Input: arr[] = {1, 3, 1, 5, 1, 3}, K = 2
Output: 4
Explanation: Splitting the array into the subsets {1, 1, 3} and {1, 3, 5} maximizes the sum of second maximum elements in the two arrays.

Input: arr[] = {1, 2, 5, 8, 6, 4, 3, 4, 9}, K = 3
Output: 17

Approach: The idea is to sort the array and keep adding every second element encountered while traversing the array in reverse, starting from the second largest element in the array, exactly K times. Follow the steps below to solve the problem:

Below is the implementation of the above approach:

## 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);` `    ``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

 ``
Output:
`4`

Time complexity: O(N logN)
Auxiliary Space: O(N)

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