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 = 2Output:4Explanation: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 = 3Output: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:

- Sort the array in ascending order.
- Traverse the array
**arr[]**in reverse. - Initialize the
**i = N – 1**. - Iterate
**K**times and add**arr[i – 1]**to the sum and reduce**i**by**2**. - Finally, print the sum obtained.

Below is the implementation of the above approach:

## C++

`// C++ program to implement` `// the above approach` `#include <bits/stdc++.h>` `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;` `}` |

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## 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.` |

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## 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` |

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## 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` |

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**Output:**

4

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

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