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
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 setsvoid 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 codeint main(){ int n = 8; find_set(n); return 0;}// This code is contributed by Sania Kumari Gupta |
C
// C implementation of the approach#include <stdio.h>// Function to find the required setsvoid find_set(int n){ // Impossible case if (n <= 2) { printf("-1"); return; } // Sum of first n-1 natural numbers int sum1 = (n * (n - 1)) / 2; int sum2 = n; printf("%d %d", sum1, sum2);}// Driver codeint main(){ int n = 8; find_set(n); return 0;}// This code is contributed by Sania Kumari Gupta |
Java
// Java implementation of the approachimport java.io.*;class GFG { // Function to find the required setsstatic 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 codepublic 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 setsdef 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 coden = 8;find_set(n);# This code is contributed by PrinciRaj1992 |
C#
// C# implementation of the approachusing System;class GFG { // Function to find the required setsstatic 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 codepublic 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)
Auxiliary Space: O(1) because it is using constant variables
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