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

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 set 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 the sets.

Examples:

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

Input: N = 2
Output: -1

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

Approach:

• If N ≤ 2 then print -1 as the only possible partition is {1} and {2} where sum of both the 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 set. So 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... `

Output:

```28 8
```

Time Complexity: O(1)

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Improved By : jit_t, vt_m, princiraj1992