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Minimum sum obtained from groups of four elements from the given array
  • Difficulty Level : Medium
  • Last Updated : 03 May, 2021

Given an array arr[] of N integers where N % 4 = 0, the task is to divide the integers into groups of four such that when the combined sum of the maximum two elements from all the groups is taken, it is minimum possible. Print the minimized sum.
Examples: 
 

Input: arr[] = {1, 1, 2, 2} 
Output:
The only group will be {1, 1, 2, 2}. 
2 + 2 = 4
Input: arr[] = {1, 1, 10, 2, 2, 2, 1, 8} 
Output: 21 
{1, 1, 2, 1} and {10, 2, 2, 8} are the groups that will 
give the minimum sum as 1 + 2 + 10 + 8 = 21. 
 

 

Approach: In order to minimize the sum, the maximum four elements from the array must be in the same group because the maximum two elements will definitely be included in the sum no matter what group they are a part of but the next two maximum elements can be prevented if they are part of this group. Making groups, in the same manner, will give the minimum sum possible. So, sort the array in descending order and starting from the first element, make groups of four consecutive elements.
Below is the implementation of the above approach: 
 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the minimum required sum
int minSum(int arr[], int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // Sort the array in descending order
    sort(arr, arr + n, greater<int>());
    for (int i = 0; i < n; i++) {
 
        // The indices which give 0 or 1 as
        // the remainder when divided by 4
        // will be the maximum two
        // elements of the group
        if (i % 4 < 2)
            sum = sum + arr[i];
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 1, 10, 2, 2, 2, 1 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << minSum(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
// Function to return the minimum required sum
static int minSum(Integer arr[], int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // Sort the array in descending order
    Arrays.sort(arr, Collections.reverseOrder());
    for (int i = 0; i < n; i++)
    {
 
        // The indices which give 0 or 1 as
        // the remainder when divided by 4
        // will be the maximum two
        // elements of the group
        if (i % 4 < 2)
            sum = sum + arr[i];
    }
 
    return sum;
}
 
// Driver code
public static void main(String[] args)
{
    Integer []arr = { 1, 1, 10, 2, 2, 2, 1 };
    int n = arr.length;
 
    System.out.println(minSum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
# Function to return the minimum required sum
def minSum(arr, n) :
 
    # To store the required sum
    sum = 0;
 
    # Sort the array in descending order
    arr.sort(reverse = True)
     
    for i in range(n) :
 
        # The indices which give 0 or 1 as
        # the remainder when divided by 4
        # will be the maximum two
        # elements of the group
        if (i % 4 < 2) :
            sum += arr[i];
     
    return sum;
 
# Driver code
if __name__ == "__main__" :
    arr = [ 1, 1, 10, 2, 2, 2, 1 ];
    n = len(arr);
    print(minSum(arr, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;
using System.Collections.Generic;            
 
class GFG
{
 
// Function to return the minimum required sum
static int minSum(int []arr, int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // Sort the array in descending order
    Array.Sort(arr);
    Array.Reverse(arr);
    for (int i = 0; i < n; i++)
    {
 
        // The indices which give 0 or 1 as
        // the remainder when divided by 4
        // will be the maximum two
        // elements of the group
        if (i % 4 < 2)
            sum = sum + arr[i];
    }
    return sum;
}
 
// Driver code
public static void Main(String[] args)
{
    int []arr = { 1, 1, 10, 2, 2, 2, 1 };
    int n = arr.Length;
 
    Console.WriteLine(minSum(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the minimum required sum
function minSum(arr, n)
{
 
    // To store the required sum
    let sum = 0;
 
    // Sort the array in descending order
    arr.sort(function(a, b){return b-a});
    for (let i = 0; i < n; i++) {
 
        // The indices which give 0 or 1 as
        // the remainder when divided by 4
        // will be the maximum two
        // elements of the group
        if (i % 4 < 2)
            sum = sum + arr[i];
    }
 
    return sum;
}
 
// Driver code
    let arr = [ 1, 1, 10, 2, 2, 2, 1 ];
    let n = arr.length;
 
    document.write(minSum(arr, n));
 
</script>
Output: 
14

 

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