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

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++

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// C++ implementation of the approach
#include <bits/stdc++.h>
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;
}

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Java














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// 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. 



















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Python3














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# 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 



















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C#














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// 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... 



















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Output:


28 8





Time Complexity: O(1)



          
          
          
          

            

    

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Improved By :  jit_t, vt_m, princiraj1992