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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 <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;
}

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




<script>
 
    // Javascript implementation of the approach
     
    // Function to find the required sets
    function find_set(n)
    {
 
        // Impossible case
        if (n <= 2)
        {
            document.write("-1");
            return;
        }
 
        // Sum of first n-1 natural numbers
        let sum1 = parseInt((n * (n - 1)) / 2, 10);
        let sum2 = n;
          document.write(sum1 + " " +sum2 + "</br>");
    }
     
    let n = 8;
    find_set(n);
 
</script>
Output: 
28 8

 

Time Complexity: O(1)
 




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