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++

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// 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;
}

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Java

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// 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

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Python 3

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# 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

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C#

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// 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 

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PHP

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<?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.
?>

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Output:

1


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Improved By : Rajput-Ji, jit_t, Ryuga