# Partition first N natural number into two sets such that their sum is not coprime

• Last Updated : 13 May, 2021

Given an integer N, the task is to partition the first N natural numbers in two non-empty sets such that the sum of these sets is not coprime to each other. If it is possible then find the possible partition then print -1 else print the sum of elements of both sets.
Examples:

Input: N = 5
Output: 10 5
{1, 2, 3, 4} and {5} are the valid partitions.
Input: N = 2
Output: -1

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Approach:

• If N ≤ 2 then print -1 as the only possible partition is {1} and {2} where the sum of both sets are coprime to each other.
• Now if N is odd then we put N in one set and first (N – 1) numbers into other sets. So the sum of the two sets will be N and N * (N – 1) / 2 and as their gcd is N, they will not be coprime to each other.
• If N is even then we do the same thing as previous and sum of the two sets will be N and N * (N – 1) / 2. As N is even, (N – 1) is not divisible by 2 but N is divisible which gives sum as (N / 2) * (N – 1) and their gcd will be N / 2. Since N / 2 is a factor of N, so they are no coprime to each other.

Below is the implementation of the above approach:

## C++

 `// C++ implementation of the approach``#include ``using` `namespace` `std;` `// Function to find the required sets``void` `find_set(``int` `n)``{` `    ``// Impossible case``    ``if` `(n <= 2) {``        ``cout << ``"-1"``;``        ``return``;``    ``}` `    ``// Sum of first n-1 natural numbers``    ``int` `sum1 = (n * (n - 1)) / 2;``    ``int` `sum2 = n;``    ``cout << sum1 << ``" "` `<< sum2;``}` `// Driver code``int` `main()``{``    ``int` `n = 8;``    ``find_set(n);` `    ``return` `0;``}`

## Java

 `// Java implementation of the approach``import` `java.io.*;` `class` `GFG``{``    ` `// Function to find the required sets``static` `void` `find_set(``int` `n)``{` `    ``// Impossible case``    ``if` `(n <= ``2``)``    ``{``        ``System.out.println (``"-1"``);``        ``return``;``    ``}` `    ``// Sum of first n-1 natural numbers``    ``int` `sum1 = (n * (n - ``1``)) / ``2``;``    ``int` `sum2 = n;``        ``System.out.println (sum1 + ``" "` `+sum2 );``}` `// Driver code``public` `static` `void` `main (String[] args)``{` `    ``int` `n = ``8``;``    ``find_set(n);``}``}` `// This code is contributed by jit_t.`

## Python3

 `# Python implementation of the approach` `# Function to find the required sets``def` `find_set(n):` `    ``# Impossible case``    ``if` `(n <``=` `2``):``        ``print``(``"-1"``);``        ``return``;` `    ``# Sum of first n-1 natural numbers``    ``sum1 ``=` `(n ``*` `(n ``-` `1``)) ``/` `2``;``    ``sum2 ``=` `n;``    ``print``(sum1, ``" "``, sum2);` `# Driver code``n ``=` `8``;``find_set(n);` `# This code is contributed by PrinciRaj1992`

## C#

 `// C# implementation of the approach``using` `System;` `class` `GFG``{``    ` `// Function to find the required sets``static` `void` `find_set(``int` `n)``{` `    ``// Impossible case``    ``if` `(n <= 2)``    ``{``        ``Console.WriteLine(``"-1"``);``        ``return``;``    ``}` `    ``// Sum of first n-1 natural numbers``    ``int` `sum1 = (n * (n - 1)) / 2;``    ``int` `sum2 = n;``        ``Console.WriteLine(sum1 + ``" "` `+sum2 );``}` `// Driver code``public` `static` `void` `Main ()``{` `    ``int` `n = 8;``    ``find_set(n);``}``}` `// This code is contributed by anuj_67...`

## Javascript

 ``
Output:
`28 8`

Time Complexity: O(1)

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