Closest sum partition (into two subsets) of numbers from 1 to n

Given an integer sequence 1, 2, 3, 4, …, n. The task is to divide it into two sets A and B in such a way that each element belongs to exactly one set and |sum(A) – sum(B)| is the minimum possible. Print the value of |sum(A) – sum(B)|.

Examples:

Input: 3 
Output: 0 
A = {1, 2} and B = {3} ans |sum(A) - sum(B)| = |3 - 3| = 0.

Input: 6 
Output: 0 
A = {1, 3, 4} and B = {2, 5} ans |sum(A) - sum(B)| = |3 - 3| = 0.

Input: 5 
Output: 1 

Approach:

Take mod = n % 4, 

1. If mod = 0 or mod = 3 then print 0.
2. If mod = 1 or mod = 2 then print 1.

This is because the groups will be chosen as A = {N, N – 3, N – 4, N – 7, N – 8, …..}, B = {N – 1, N – 2, N – 5, N – 6, …..} 
Starting from N to 1, place 1st element in group A then alternate every 2 elements in B, A, B, A, …..  

• When n % 4 = 0: N = 8, A = {1, 4, 5, 8} and B = {2, 3, 6, 7}
• When n % 4 = 1: N = 9, A = {1, 4, 5, 8, 9} and B = {2, 3, 6, 7}
• When n % 4 = 2: N = 10, A = {1, 4, 5, 8, 9} and B = {2, 3, 6, 7, 10}
• When n % 4 = 3: N = 11, A = {1, 4, 5, 8, 9} and B = {2, 3, 6, 7, 10, 11}

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
// absolute difference
int minAbsDiff(int n)
{
    int mod = n % 4;

    if (mod == 0 || mod == 3)
        return 0;

    return 1;
}

// Driver code
int main()
{
    int n = 5;
    cout << minAbsDiff(n);

    return 0;
}

Java

// Java implementation of the approach
class GFG
{

// Function to return the minimum required
// absolute difference

    static int minAbsDiff(int n)
    {
        int mod = n % 4;
        if (mod == 0 || mod == 3)
        {
            return 0;
        }
        return 1;
    }

    // Driver code
    public static void main(String[] args)
    {
        int n = 5;
        System.out.println(minAbsDiff(n));
    }
}

// This code is contributed by Rajput-JI

Python 3

# Python3 implementation of the approach

# Function to return the minimum required
# absolute difference
def minAbsDiff(n) :
    mod = n % 4;

    if (mod == 0 or mod == 3) :
        return 0;

    return 1;

# Driver code
if __name__ == "__main__" :

    n = 5;
    print(minAbsDiff(n))

# This code is contributed by Ryuga

C#

// C# implementation of the
// above approach
using System;

class GFG
{

    // Function to return the minimum
    // required absolute difference
    static int minAbsDiff(int n)
    {
        int mod = n % 4;
        if (mod == 0 || mod == 3)
        {
            return 0;
        }
        return 1;
    }

    // Driver code
    static public void Main ()
    {
        int n = 5;
        Console.WriteLine(minAbsDiff(n));
    }
}

// This code is contributed by akt_mit

PHP

<?php
// PHP implementation of the approach

// Function to return the minimum
// required absolute difference
function minAbsDiff($n)
{
    $mod = $n % 4;

    if ($mod == 0 || $mod == 3)
        return 0;

    return 1;
}

// Driver code
$n = 5;
echo minAbsDiff($n);

// This code is contributed by Tushil.
?>

Javascript

<script>
    // Javascript implementation of the above approach

    // Function to return the minimum
    // required absolute difference
    function minAbsDiff(n)
    {
        let mod = n % 4;
        if (mod == 0 || mod == 3)
        {
            return 0;
        }
        return 1;
    }

    let n = 5;
      document.write(minAbsDiff(n));

</script>

Output:

1

Time Complexity: O(1) // since no loop is used so the algorithm takes constant time to execute completely.

Auxiliary Space: O(1) // since no extra array is used the algorithm takes up constant space to run completely.