# 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:3Output:0

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

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

**Approach:** Take **mod = n % 4**,

- If
**mod = 0**or**mod = 3**then print**0**. - 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