# Split array into K subsets to maximize their sum of maximums and minimums

Given an integer **K** and an array **A[ ]** whose length is multiple of **K**, the task is to split the elements of the given array into **K** subsets, each having an equal number of elements, such that the sum of the maximum and minimum elements of each subset is the maximum summation possible.

**Examples:**

Input:K = 2, A[ ] = {1, 13, 7, 17, 6, 5}Output:37Explanation:

1st group: {1, 5, 17} maximum = 17, minimum = 1

2nd group: {6, 7, 13} maximum = 13, minimum = 6

Hence, maximum possible sum = 17 + 1 + 13 + 6 = 37Input:K = 2, A[ ] = {10, 10, 10, 10, 11, 11}Output:42Explanation:

1st group: {11, 10, 10} maximum = 11, minimum = 10

2nd group: {11, 10, 10} maximum = 11, minimum = 10

Hence, maximum sum possible = 11 + 10 + 11 + 10 = 42

**Naive Approach:**

The simplest approach to solve this problem is to generate all possible groups of **K** subsets of size **N/K** and for each group, find maximum and minimum in every subset and calculate their sum. Once the sum of all groups is calculated, print the maximum sum obtained. **Time Complexity:** O(2^{N})**Auxiliary Space:** O(N)**Efficient Approach:**

The idea is to optimize the above approach using the Greedy Technique. Since the maximum sum of the maximum and minimum element from each subset is needed, try to maximize the maximum element and minimum element. For the maximum element of each subset, take first **K** largest elements from the given array and insert each one to different subsets. For the minimum element of each subset, from the sorted array, starting from index **0**, pick every next element at **(N / K) – 1** interval since the size of each subset is **N / K** and each one already contains a maximum element.

Follow the steps below:

- Calculate the number of elements in each group i.e.
**(N/K)**. - Sort all the elements of
**A[ ]**in non-descending order. - For the sum of
**maximum**elements, add all**K**largest elements from the sorted array. - For the sum of
**minimum**elements, starting from index**0**, select**K**elements each with**(N / K) – 1**interval and add them. - Finally, calculate the sum of maximum and the sum of minimum elements. Print the sum of their respective sums as the final answer.

Below is the implementation of the above approach:

## C++

`// C++ Program to implement` `// the above approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function that prints` `// the maximum sum possible` `void` `maximumSum(` `int` `arr[],` ` ` `int` `n, ` `int` `k)` `{` ` ` `// Find elements in each group` ` ` `int` `elt = n / k;` ` ` `int` `sum = 0;` ` ` `// Sort all elements in` ` ` `// non-descending order` ` ` `sort(arr, arr + n);` ` ` `int` `count = 0;` ` ` `int` `i = n - 1;` ` ` `// Add K largest elements` ` ` `while` `(count < k) {` ` ` `sum += arr[i];` ` ` `i--;` ` ` `count++;` ` ` `}` ` ` `count = 0;` ` ` `i = 0;` ` ` `// For sum of minimum` ` ` `// elements from each subset` ` ` `while` `(count < k) {` ` ` `sum += arr[i];` ` ` `i += elt - 1;` ` ` `count++;` ` ` `}` ` ` `// Printing the maximum sum` ` ` `cout << sum << ` `"\n"` `;` `}` `// Driver Code` `int` `main()` `{` ` ` `int` `Arr[] = { 1, 13, 7, 17, 6, 5 };` ` ` `int` `K = 2;` ` ` `int` `size = ` `sizeof` `(Arr) / ` `sizeof` `(Arr[0]);` ` ` `maximumSum(Arr, size, K);` ` ` `return` `0;` `}` |

## Java

`// Java program to implement` `// the above approach` `import` `java.util.Arrays;` `class` `GFG{` ` ` `// Function that prints` `// the maximum sum possible` `static` `void` `maximumSum(` `int` `arr[],` ` ` `int` `n, ` `int` `k)` `{` ` ` ` ` `// Find elements in each group` ` ` `int` `elt = n / k;` ` ` `int` `sum = ` `0` `;` ` ` `// Sort all elements in` ` ` `// non-descending order` ` ` `Arrays.sort(arr);` ` ` `int` `count = ` `0` `;` ` ` `int` `i = n - ` `1` `;` ` ` `// Add K largest elements` ` ` `while` `(count < k)` ` ` `{` ` ` `sum += arr[i];` ` ` `i--;` ` ` `count++;` ` ` `}` ` ` `count = ` `0` `;` ` ` `i = ` `0` `;` ` ` `// For sum of minimum` ` ` `// elements from each subset` ` ` `while` `(count < k)` ` ` `{` ` ` `sum += arr[i];` ` ` `i += elt - ` `1` `;` ` ` `count++;` ` ` `}` ` ` `// Printing the maximum sum` ` ` `System.out.println(sum);` `}` `// Driver code` `public` `static` `void` `main (String[] args)` `{` ` ` `int` `Arr[] = { ` `1` `, ` `13` `, ` `7` `, ` `17` `, ` `6` `, ` `5` `};` ` ` `int` `K = ` `2` `;` ` ` `int` `size = Arr.length;` ` ` `maximumSum(Arr, size, K);` `}` `}` `// This code is contributed by Shubham Prakash` |

## Python3

`# Python3 program to implement` `# the above approach` `# Function that prints` `# the maximum sum possible` `def` `maximumSum(arr, n, k):` ` ` ` ` `# Find elements in each group` ` ` `elt ` `=` `n ` `/` `/` `k;` ` ` `sum` `=` `0` `;` ` ` `# Sort all elements in` ` ` `# non-descending order` ` ` `arr.sort();` ` ` `count ` `=` `0` `;` ` ` `i ` `=` `n ` `-` `1` `;` ` ` `# Add K largest elements` ` ` `while` `(count < k):` ` ` `sum` `+` `=` `arr[i];` ` ` `i ` `-` `=` `1` `;` ` ` `count ` `+` `=` `1` `;` ` ` `count ` `=` `0` `;` ` ` `i ` `=` `0` `;` ` ` `# For sum of minimum` ` ` `# elements from each subset` ` ` `while` `(count < k):` ` ` `sum` `+` `=` `arr[i];` ` ` `i ` `+` `=` `elt ` `-` `1` `;` ` ` `count ` `+` `=` `1` `;` ` ` `# Printing the maximum sum` ` ` `print` `(` `sum` `);` `# Driver code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `Arr ` `=` `[` `1` `, ` `13` `, ` `7` `, ` `17` `, ` `6` `, ` `5` `];` ` ` `K ` `=` `2` `;` ` ` `size ` `=` `len` `(Arr);` ` ` `maximumSum(Arr, size, K);` `# This code is contributed by sapnasingh4991` |

## C#

`// C# program to implement` `// the above approach` `using` `System;` `class` `GFG{` ` ` `// Function that prints` `// the maximum sum possible` `static` `void` `maximumSum(` `int` `[]arr,` ` ` `int` `n, ` `int` `k)` `{` ` ` ` ` `// Find elements in each group` ` ` `int` `elt = n / k;` ` ` `int` `sum = 0;` ` ` `// Sort all elements in` ` ` `// non-descending order` ` ` `Array.Sort(arr);` ` ` `int` `count = 0;` ` ` `int` `i = n - 1;` ` ` `// Add K largest elements` ` ` `while` `(count < k)` ` ` `{` ` ` `sum += arr[i];` ` ` `i--;` ` ` `count++;` ` ` `}` ` ` `count = 0;` ` ` `i = 0;` ` ` `// For sum of minimum` ` ` `// elements from each subset` ` ` `while` `(count < k)` ` ` `{` ` ` `sum += arr[i];` ` ` `i += elt - 1;` ` ` `count++;` ` ` `}` ` ` `// Printing the maximum sum` ` ` `Console.WriteLine(sum);` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `[]Arr = { 1, 13, 7, 17, 6, 5 };` ` ` `int` `K = 2;` ` ` `int` `size = Arr.Length;` ` ` `maximumSum(Arr, size, K);` `}` `}` `// This code is contributed by amal kumar choubey` |

## Javascript

`<script>` `// Javascript program implementation` `// of the approach` `// Function that prlets` `// the maximum sum possible` `function` `maximumSum(arr, n, k)` `{` ` ` ` ` `// Find elements in each group` ` ` `let elt = (n / k);` ` ` ` ` `let sum = 0;` ` ` ` ` `// Sort all elements in` ` ` `// non-descending order` ` ` `arr.sort((a, b) => a - b);` ` ` ` ` `let count = 0;` ` ` `let i = n - 1;` ` ` ` ` `// Add K largest elements` ` ` `while` `(count < k)` ` ` `{` ` ` `sum += arr[i];` ` ` `i--;` ` ` `count++;` ` ` `}` ` ` `count = 0;` ` ` `i = 0;` ` ` ` ` `// For sum of minimum` ` ` `// elements from each subset` ` ` `while` `(count < k)` ` ` `{` ` ` `sum += arr[i];` ` ` `i += elt - 1;` ` ` `count++;` ` ` `}` ` ` ` ` `// Prleting the maximum sum` ` ` `document.write(sum);` `}` `// Driver Code` ` ` ` ` `let Arr = [ 1, 13, 7, 17, 6, 5 ];` ` ` ` ` `let K = 2;` ` ` ` ` `let size = Arr.length;` ` ` ` ` `maximumSum(Arr, size, K);` ` ` `</script>` |

**Output:**

37

**Time complexity:** O(N*logN)**Auxiliary Space:** O(1)

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