Maximize the sum of Array by formed by adding pair of elements

Given an array a[] of 2*N integers, The task is to make the array a[] of size N i.e, reducing it to half size such that, a[i] = ⌊(a[j] + a[k]) / N⌋,  0 < j, k < 2*N – 1. and form the array so that the sum of all the elements of the array a[], will be maximum. Output the maximum sum.

Examples:

Input: arr[] = {3, 5, 10, 8, 4, 7}, N = 3
Output: 12
Explanation: If we form, a[] = {4 + 5, 7+8, 3+10} = {9, 15, 13}, 
Sum = floor(9/3) + floor(15/3) + floor(13/3) = 3 + 5 + 4 = 12.

Input: arr[] = {1, 2}, N = 1
Output: 3

Approach: To solve the problem follow the below idea:

The idea is to pair the elements whose sum (a[i]+a[j])%N > a[i]%N + a[j]%N, for this we have to store a[i]/N for 0<i<2*N-1 and modify a[i] = a[i]%N for finding the remainder. Now we have to from i with j such that a[i] + a[j] >= N [0<(a[i]+a[j])<2N-2], because, we have to take as many extra count that we were losing with floor division. 

For this we have to sort the array and initialize pointers i = 2*N-1and  j=0, and start forming pairs with last element because it is the greatest, if it cannot form pair with sum ≥ N then any other pair does not form, we have to from the pairs of available largest element with smallest element such that sum ≥ N if possible.

Follow the below steps to solve the problem:

• First of all, we have to store the sum of the given array by sum=sum+arr[i]/n and update the value of the array by arr[i]=arr[i]%n. 
• After that, sort the array.
• Then, by using the two-pointers approach, find the pairs with a sum greater than or equal to n.
• Then return the sum.

Below is the implementation of the above approach: 

C++

// C++ code for above approach

#include <bits/stdc++.h>
using namespace std;

// Function to form the array and return
// maximum possible sum
int solve(int* a, int n)
{

    // Initialize result
    int Sum = 0;

    for (int i = 0; i < 2 * n; i++) {
        Sum = Sum + a[i] / n;
        a[i] = a[i] % n;
    }

    // Sort the array
    sort(a, a + 2 * n);

    // Initialize pointers
    int i = 2 * n - 1;
    int j = 0;

    // Find pairs with sum greater
    // than or equal to N
    while (i > j) {
        if (a[i] + a[j] >= n) {
            Sum++;
            i--;
            j++;
        }
        else {
            j++;
        }
    }

    // Return maximum possible sum
    return Sum;
}

// Driver Code
int main()
{

    int arr[] = { 3, 5, 10, 8, 4, 7 };

    int N = 3;

    // Function Call
    cout << solve(arr, N) << endl;

    return 0;
}

Java

// Java code for above approach

import java.io.*;
import java.util.*;

class GFG {

    // Function to form the array and return
    // maximum possible sum
    static int solve(int[] a, int n)
    {
        // Initialize result
        int Sum = 0;

        for (int i = 0; i < 2 * n; i++) {
            Sum = Sum + a[i] / n;
            a[i] = a[i] % n;
        }

        // Sort the array
        Arrays.sort(a);

        // Initialize pointers
        int i = 2 * n - 1;
        int j = 0;

        // Find pairs with sum greater
        // than or equal to N
        while (i > j) {
            if (a[i] + a[j] >= n) {
                Sum++;
                i--;
                j++;
            }
            else {
                j++;
            }
        }

        // Return maximum possible sum
        return Sum;
    }

    public static void main(String[] args)
    {
        int[] arr = { 3, 5, 10, 8, 4, 7 };
        int N = 3;

        // Function call
        System.out.println(solve(arr, N));
    }
}

// This code is contributed by lokeshmvs21.

Python3

# Python code for above approach

# Function to form the array and return
# maximum possible sum
def solve(a, n):
    # Initialize result
    Sum = 0
    
    for i in range(2 * n):
        Sum = Sum + int(a[i] / n)
        a[i] = a[i] % n
        
    # Sort the array
    a.sort()
    
    # Initialize pointers
    i = 2 * n - 1
    j = 0
    
    # Find pairs with sum greater
    # than or equal to N
    while i > j:
        if a[i] + a[j] >= n:
            Sum += 1
            i -= 1
            j += 1
        else:
            j += 1
            
    # Return maximum possible sum
    return Sum

# Driver Code
arr = [3, 5, 10, 8, 4, 7]
N = 3

# Function call
print(solve(arr, N))

# This code is contributed by Tapesh(tapeshdua420)

C#

// C# code
using System;

public class GFG {

  // Function to form the array and return
  // maximum possible sum
  public static int solve(int[] a, int n)
  {

    // Initialize result
    int Sum = 0;

    for (int k = 0; k < 2 * n; k++) {
      Sum = Sum + (int)(a[k] / n);
      a[k] = a[k] % n;
    }

    // Sort the array
    Array.Sort(a, 0, 2 * n);

    // Initialize pointers
    int i = 2 * n - 1;
    int j = 0;

    // Find pairs with sum greater
    // than or equal to N
    while (i > j) {
      if (a[i] + a[j] >= n) {
        Sum++;
        i--;
        j++;
      }
      else {
        j++;
      }
    }

    // Return maximum possible sum
    return Sum;
  }

  static public void Main()
  {

    int[] arr = { 3, 5, 10, 8, 4, 7 };

    int N = 3;

    // Function Call
    Console.WriteLine(solve(arr, N));
  }
}

// This code is contributed by ksam24000.

Javascript

  // Javascript code for above approach

  // Function to form the array and return
  // maximum possible sum
  function solve(a, n) {
      // Initialize result
      var Sum = 0;

      for (var i = 0; i < 2 * n; i++) {
          Sum += parseInt(a[i] / n);
          a[i] = a[i] % n;
      }

      // Sort the array
      a.sort();

      // Initialize pointers
      var i = 2 * n - 1;
      var j = 0;

      // Find pairs with sum greater
      // than or equal to N
      while (i > j) {
          if (a[i] + a[j] >= n) {
              Sum++;
              i--;
              j++;
          } else {
              j++;
          }
      }

      // Return maximum possible sum
      return Sum;
  }

  // Driver Code
  var arr = [3, 5, 10, 8, 4, 7];
  var N = 3;

  // Function call
  console.log(solve(arr, N));

  // This code is contributed by Tapesh(tapeshdua420)

Output

12

Time Complexity: O(N * log N)
Auxiliary Space: O(1)

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