Minimize the absolute difference of sum of two subsets

Given a number n, divide first n natural numbers (1, 2, …n) into two subsets such that difference between sums of two subsets is minimum.
Examples: 

Input : n = 4
Output : First subset sum = 5, 
         Second subset sum = 5.
         Difference = 0
Explanation:
Subset 1: 1 4 
Subset 2: 2 3 

Input : n = 6 
Output: First subset sum = 10, 
        Second subset sum = 11.
        Difference = 1
Explanation : 
Subset 1: 1 3 6 
Subset 2: 2 4 5 

Approach:
The approach is based on the fact that any four consecutive numbers can be divided into two groups by putting middle two elements in one group and corner elements in other group. So, if n is a multiple of 4 then their difference will be 0, hence the summation of one set will be half of the summation of N elements which can be calculated by using sum = n*(n+1)/2
There are three other cases to consider in which we cannot divide into groups of 4, which will leave a remainder of 1, 2 and 3: 
a) If it leaves a remainder of 1, then all other n-1 elements are clubbed into group of 4 hence their sum will be int(sum/2) and the other half sum will be int(sum/2+1) and their difference will always be 1. 
b) Above mentioned steps will be followed in case of n%4 == 2 also. Here we form groups of size 4 for elements from 3 onward. Remaining elements would be 1 and 2. 1 goes in one group and 2 goes in other group. 
c) When n%4 == 3, then club n-3 elements into groups of 4. The left out elements will be 1, 2 and 3, in which 1 and 2 will go to one set and 3 to the other set which eventually makes the difference to be 0 and summation of each set to be sum/2.
Below is the implementation of the above approach: 

CPP

// CPP program to Minimize the absolute
// difference of sum of two subsets
#include <bits/stdc++.h>
using namespace std;

// function to print difference
void subsetDifference(int n)
{
    // summation of n elements
    int s = n * (n + 1) / 2;

    // if divisible by 4
    if (n % 4 == 0) {
        cout << "First subset sum = "
             << s / 2;
        cout << "\nSecond subset sum = "
             << s / 2;
        cout << "\nDifference = " << 0;
    }
    else {

        // if remainder 1 or 2. In case of remainder
        // 2, we divide elements from 3 to n in groups
        // of size 4 and put 1 in one group and 2 in
        // group. This also makes difference 1.
        if (n % 4 == 1 || n % 4 == 2) {

            cout << "First subset sum = "
                 << s / 2;
            cout << "\nSecond subset sum = "
                 << s / 2 + 1;
            cout << "\nDifference = " << 1;
        }

        // We put elements from 4 to n in groups of
        // size 4. Remaining elements 1, 2 and 3 can
        // be divided as (1, 2) and (3).
        else
        {
            cout << "First subset sum = "
                 << s / 2;
            cout << "\nSecond subset sum = "
                 << s / 2;
            cout << "\nDifference = " << 0;
        }
    }
}

// driver program to test the above function
int main()
{
    int n = 6;
    subsetDifference(n);
    return 0;
}

Java

// Java program for Minimize the absolute
// difference of sum of two subsets
import java.util.*;

class GFG {

    // function to print difference
    static void subsetDifference(int n)
    {
        // summation of n elements
        int s = n * (n + 1) / 2;

        // if divisible by 4
        if (n % 4 == 0) {

                 System.out.println("First subset sum = " + s / 2);
                 System.out.println("Second subset sum = " + s / 2);
                 System.out.println("Difference = " + 0);
        }
        else {

            // if remainder 1 or 2. In case of remainder
            // 2, we divide elements from 3 to n in groups
            // of size 4 and put 1 in one group and 2 in
            // group. This also makes difference 1.
            if (n % 4 == 1 || n % 4 == 2) {

                  System.out.println("First subset sum = " + s / 2);
                  System.out.println("Second subset sum = " + ((s / 2) + 1));
                  System.out.println("Difference = " + 1);
            }

            // We put elements from 4 to n in groups of
            // size 4. Remaining elements 1, 2 and 3 can
            // be divided as (1, 2) and (3).
            else
            {
                 System.out.println("First subset sum = " + s / 2);
                 System.out.println("Second subset sum = " + s / 2);
                 System.out.println("Difference = " + 0);
            }
        }
    }

    /* Driver program to test above function */
    public static void main(String[] args)
    {
         int n = 6;
         subsetDifference(n);
    }
}

// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 code to Minimize the absolute
# difference of sum of two subsets

# function to print difference
def subsetDifference( n ):

    # summation of n elements
    s = int(n * (n + 1) / 2)

    # if divisible by 4
    if n % 4 == 0:
        print("First subset sum = ", int(s / 2))
        print("Second subset sum = ",int(s / 2))
        print("Difference = " , 0)

    else:

        # if remainder 1 or 2. In case of remainder
        # 2, we divide elements from 3 to n in groups
        # of size 4 and put 1 in one group and 2 in
        # group. This also makes difference 1.
        if n % 4 == 1 or n % 4 == 2:
            print("First subset sum = ",int(s/2))
            print("Second subset sum = ",int(s/2)+1)
            print("Difference = ", 1)

        # We put elements from 4 to n in groups of
        # size 4. Remaining elements 1, 2 and 3 can
        # be divided as (1, 2) and (3).
        else:
            print("First subset sum = ", int(s / 2))
            print("Second subset sum = ",int(s / 2))
            print("Difference = " , 0)

# driver code to test the above function
n = 6
subsetDifference(n)

# This code is contributed by "Sharad_Bhardwaj".

C#

// C# program for Minimize the absolute
// difference of sum of two subsets
using System;

class GFG {

    // function to print difference
    static void subsetDifference(int n)
    {

        // summation of n elements
        int s = n * (n + 1) / 2;

        // if divisible by 4
        if (n % 4 == 0) {

            Console.WriteLine("First "
            + "subset sum = " + s / 2);

            Console.WriteLine("Second "
            + "subset sum = " + s / 2);

            Console.WriteLine("Difference"
                              + " = " + 0);
        }
        else {

            // if remainder 1 or 2. In case
            // of remainder 2, we divide
            // elements from 3 to n in groups
            // of size 4 and put 1 in one
            // group and 2 in group. This
            // also makes difference 1.
            if (n % 4 == 1 || n % 4 == 2) {

                Console.WriteLine("First "
                + "subset sum = " + s / 2);

                Console.WriteLine("Second "
                + "subset sum = " + ((s / 2)
                                      + 1));

                Console.WriteLine("Difference"
                               + " = " + 1);
            }

            // We put elements from 4 to n
            // in groups of size 4. Remaining
            // elements 1, 2 and 3 can
            // be divided as (1, 2) and (3).
            else
            {
                Console.WriteLine("First "
                + "subset sum = " + s / 2);

                Console.WriteLine("Second "
                 + "subset sum = " + s / 2);

                Console.WriteLine("Difference"
                                + " = " + 0);
            }
        }
    }

    /* Driver program to test above
    function */
    public static void Main()
    {
        int n = 6;

        subsetDifference(n);
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP program to Minimize the absolute
// difference of sum of two subsets

// function to print difference
function subsetDifference($n)
{

    // summation of n elements
    $s = $n * ($n + 1) / 2;

    // if divisible by 4
    if ($n % 4 == 0)
    {
        echo "First subset sum = "
            ,floor($s / 2);
        echo "\nSecond subset sum = "
            ,floor($s / 2);
        echo "\nDifference = " , 0;
    }
    else
    {

        // if remainder 1 or 2.
        // In case of remainder
        // 2, we divide elements
        // from 3 to n in groups
        // of size 4 and put 1 in
        // one group and 2 in
        // group. This also makes
        // difference 1.
        if ($n % 4 == 1 || $n % 4 == 2)
        {

            echo"First subset sum = "
                , floor($s / 2);
            echo "\nSecond subset sum = "
                , floor($s / 2 + 1);
            echo"\nDifference = " ,1;
        }

        // We put elements from 4
        // to n in groups of
        // size 4. Remaining
        // elements 1, 2 and 3 can
        // be divided as (1, 2)
        // and (3).
        else
        {
            echo "First subset sum = "
                ,floor($s / 2);
            echo "\nSecond subset sum = "
                , floor($s / 2);
            echo"\nDifference = " , 0;
        }
    }
}

    // Driver code
    $n = 6;
    subsetDifference($n);

// This code is contributed by anuj_67.
?>

Javascript

<script>
// Javascript program to Minimize the absolute
// difference of sum of two subsets

// function to print difference
function subsetDifference(n)
{

    // summation of n elements
    let s = n * (n + 1) / 2;

    // if divisible by 4
    if (n % 4 == 0)
    {
        document.write("First subset sum = " + Math.floor(s / 2));
        document.write("<br> Second subset sum = " + Math.floor(s / 2));
        document.write("<br> Difference = "  +  0);
    }
    else
    {

        // if remainder 1 or 2.
        // In case of remainder
        // 2, we divide elements
        // from 3 to n in groups
        // of size 4 and put 1 in
        // one group and 2 in
        // group. This also makes
        // difference 1.
        if (n % 4 == 1 || n % 4 == 2)
        {

            document.write("First subset sum = " + Math.floor(s / 2));
            document.write("<br> Second subset sum = " + Math.floor(s / 2 + 1));
            document.write("<br> Difference = " + 1);
        }

        // We put elements from 4
        // to n in groups of
        // size 4. Remaining
        // elements 1, 2 and 3 can
        // be divided as (1, 2)
        // and (3).
        else
        {
            document.write( "First subset sum = " + Math.floor(s / 2));
            document.write( "<br> Second subset sum = " + Math.floor(s / 2));
            document.write("<br> Difference = " + 0);
        }
    }
}

    // Driver code
    let n = 6;
    subsetDifference(n);

// This code is contributed by _saurabh_jaiswal.
</script>

Output

First subset sum = 10
Second subset sum = 11
Difference = 1

Time Complexity: O(1)
Auxiliary Space: O(1)