# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

Output:

```1
```

Don’t stop now and take your learning to the next level. Learn all the important concepts of Data Structures and Algorithms with the help of the most trusted course: DSA Self Paced. Become industry ready at a student-friendly price.

My Personal Notes arrow_drop_up Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.

Improved By : Rajput-Ji, jit_t, AnkitRai01