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++
// 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; } |
chevron_right
filter_none
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. |
chevron_right
filter_none
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 |
chevron_right
filter_none
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... |
chevron_right
filter_none
Output:
28 8
Time Complexity: O(1)
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
