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: 37
Explanation:
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 = 37
Input: K = 2, A[ ] = {10, 10, 10, 10, 11, 11}
Output: 42
Explanation:
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(2N)
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++
#include <bits/stdc++.h>
using namespace std;
void maximumSum( int arr[],
int n, int k)
{
int elt = n / k;
int sum = 0;
sort(arr, arr + n);
int count = 0;
int i = n - 1;
while (count < k) {
sum += arr[i];
i--;
count++;
}
count = 0;
i = 0;
while (count < k) {
sum += arr[i];
i += elt - 1;
count++;
}
cout << sum << "\n" ;
}
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
import java.util.Arrays;
class GFG{
static void maximumSum( int arr[],
int n, int k)
{
int elt = n / k;
int sum = 0 ;
Arrays.sort(arr);
int count = 0 ;
int i = n - 1 ;
while (count < k)
{
sum += arr[i];
i--;
count++;
}
count = 0 ;
i = 0 ;
while (count < k)
{
sum += arr[i];
i += elt - 1 ;
count++;
}
System.out.println(sum);
}
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);
}
}
|
Python3
def maximumSum(arr, n, k):
elt = n / / k;
sum = 0 ;
arr.sort();
count = 0 ;
i = n - 1 ;
while (count < k):
sum + = arr[i];
i - = 1 ;
count + = 1 ;
count = 0 ;
i = 0 ;
while (count < k):
sum + = arr[i];
i + = elt - 1 ;
count + = 1 ;
print ( sum );
if __name__ = = '__main__' :
Arr = [ 1 , 13 , 7 , 17 , 6 , 5 ];
K = 2 ;
size = len (Arr);
maximumSum(Arr, size, K);
|
C#
using System;
class GFG{
static void maximumSum( int []arr,
int n, int k)
{
int elt = n / k;
int sum = 0;
Array.Sort(arr);
int count = 0;
int i = n - 1;
while (count < k)
{
sum += arr[i];
i--;
count++;
}
count = 0;
i = 0;
while (count < k)
{
sum += arr[i];
i += elt - 1;
count++;
}
Console.WriteLine(sum);
}
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);
}
}
|
Javascript
<script>
function maximumSum(arr, n, k)
{
let elt = (n / k);
let sum = 0;
arr.sort((a, b) => a - b);
let count = 0;
let i = n - 1;
while (count < k)
{
sum += arr[i];
i--;
count++;
}
count = 0;
i = 0;
while (count < k)
{
sum += arr[i];
i += elt - 1;
count++;
}
document.write(sum);
}
let Arr = [ 1, 13, 7, 17, 6, 5 ];
let K = 2;
let size = Arr.length;
maximumSum(Arr, size, K);
</script>
|
Time complexity: O(N*logN)
Auxiliary Space: O(1)
Last Updated :
24 Nov, 2021
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