Given an integer N, the task is to create two sets of distinct elements from 1 to N such that the gcd of their respective sums is greater than 1. Print the respective sets. If no such split can be done, print -1.
Examples:
Input: N = 5
Output:
2 4
1 3 5
Explanation:
GCD(Sum({2, 4}), Sum({1, 3, 5}) = GCD(6, 9) = 3Input: N = 2
Output: -1
Explanation:
For N = 2, it is not possible to divide into two sets having GCD of their sum greater than 1.
Approach 1: A simple approach is to split all even numbers up to N into one set and all odd numbers into another set.
Below code is the implementation of the approach:
C++
// C++ program to split N natural numbers // into two sets having GCD // of their sums greater than 1 #include <bits/stdc++.h> using namespace std;
// Function to create // and print the two sets void createSets(int N)
{ // No such split
// possible for N <= 2
if (N <= 2) {
cout << "-1" << endl;
return;
}
// Print the first set
// consisting of even elements
for (int i = 2; i <= N; i += 2)
cout << i << " ";
cout << "\n";
// Print the second set
// consisting of odd ones
for (int i = 1; i <= N; i += 2) {
cout << i << " ";
}
} // Driver Code int main()
{ int N = 6;
createSets(N);
return 0;
} |
Java
// Java program to split N natural numbers // into two sets having GCD // of their sums greater than 1 class GFG{
// Function to create // and print the two sets static void createSets(int N)
{ // No such split
// possible for N <= 2
if (N <= 2)
{
System.out.println("-1" );
return;
}
// Print the first set
// consisting of even elements
for (int i = 2; i <= N; i += 2)
System.out.print(i + " ");
System.out.print("\n") ;
// Print the second set
// consisting of odd ones
for (int i = 1; i <= N; i += 2)
{
System.out.print(i + " ");
}
} // Driver Code public static void main(String[] args)
{ int N = 6;
createSets(N);
} } // This code is contributed by shivanisinghss2110 |
Python3
# Python3 program to split N natural numbers # into two sets having GCD # of their sums greater than 1 # Function to create # and print the two sets def createSets(N):
# No such split
# possible for N <= 2
if (N <= 2):
print("-1");
return;
# Print the first set
# consisting of even elements
for i in range(2, N + 1, 2):
print(i, end=" ");
print("");
# Print the second set
# consisting of odd ones
for i in range(1, N + 1, 2):
print(i, end = " ");
# Driver Code if __name__ == '__main__':
N = 6;
createSets(N);
# This code is contributed by gauravrajput1 |
C#
// C# program to split N natural numbers // into two sets having GCD // of their sums greater than 1 using System;
class GFG{
// Function to create // and print the two sets static void createSets(int N)
{ // No such split
// possible for N <= 2
if (N <= 2)
{
Console.WriteLine("-1" );
return;
}
// Print the first set
// consisting of even elements
for (int i = 2; i <= N; i += 2)
Console.Write(i + " ");
Console.Write("\n") ;
// Print the second set
// consisting of odd ones
for (int i = 1; i <= N; i += 2)
{
Console.Write(i + " ");
}
} // Driver Code public static void Main(String[] args)
{ int N = 6;
createSets(N);
} } // This code is contributed by shivanisinghss2110 |
Javascript
<script> // Javascript program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
// Function to create
// and print the two sets
function createSets(N)
{
// No such split
// possible for N <= 2
if (N <= 2) {
document.write("-1");
return;
}
// Print the first set
// consisting of even elements
for (let i = 2; i <= N; i += 2)
document.write(i + " ");
document.write("</br>");
// Print the second set
// consisting of odd ones
for (let i = 1; i <= N; i += 2) {
document.write(i + " ");
}
}
let N = 6;
createSets(N);
</script> |
Output:
2 4 6 1 3 5
Time Complexity: O(N)
Auxiliary Space: O(1),since no extra space has been taken.
Approach 2: This approach can be divided into 2 cases based on N:
-
If N is odd:
- The sum of N natural numbers is divisible by (N+1)/2 in this case.
- Therefore we only need to put (N+1)/2 into first set and the remaining numbers into second set.
- This will automatically make the GCD of their sums greater than 1.
-
If N is even:
- The sum of N natural numbers is divisible by N/2 in this case.
- Therefore we only need to put N/2 into first set and the remaining numbers into second set.
- This will automatically make the GCD of their sums greater than 1.
Below is the implementation of the above approach:
C++
// C++ program to split N natural numbers // into two sets having GCD // of their sums greater than 1 #include <bits/stdc++.h> using namespace std;
#define ll long long // Function to create // and print the two sets void createSets(ll n)
{ // For n <= 2 such sets
// can never be formed
if (n <= 2) {
cout << "-1";
return;
}
else {
// Check if N is even or odd
// and store the element
// which divides the sum of N
// natural numbers accordingly
ll x = (n % 2 == 0) ? (n / 2)
: ((n + 1) / 2);
// First set
cout << x << endl;
// Print elements of second set
for (ll i = 1; i <= n; i++) {
if (i == x)
continue;
cout << i << " ";
}
}
return;
} // Driver code int main()
{ ll N = 7;
createSets(N);
return 0;
} |
Java
// Java program to split N natural numbers // into two sets having GCD // of their sums greater than 1 class GFG{
// Function to create // and print the two sets static void createSets(long n)
{ // For n <= 2 such sets
// can never be formed
if (n <= 2)
{
System.out.print("-1");
return;
}
else
{
// Check if N is even or odd
// and store the element
// which divides the sum of N
// natural numbers accordingly
long x = (n % 2 == 0) ? (n / 2) :
((n + 1) / 2);
// First set
System.out.print(x + "\n");
// Print elements of second set
for(int i = 1; i <= n; i++)
{
if (i == x)
continue;
System.out.print(i + " ");
}
}
return;
} // Driver code public static void main(String[] args)
{ long N = 7;
createSets(N);
} } // This code is contributed by PrinciRaj1992 |
Python3
# Python3 program to split N natural numbers # into two sets having GCD # of their sums greater than 1 # Function to create # and print the two sets def createSets(n):
# For n <= 2 such sets
# can never be formed
if (n <= 2):
print("-1");
return;
else:
# Check if N is even or odd
# and store the element
# which divides the sum of N
# natural numbers accordingly
x = (n // 2) if(n % 2 == 0) else (
(n + 1) // 2);
# First set
print(x, " ");
# Print elements of second set
for i in range(1, n + 1):
if (i == x):
continue;
print(i, end = " ");
return;
# Driver code if __name__ == '__main__':
N = 7;
createSets(N);
# This code is contributed by 29AjayKumar |
C#
// C# program to split N natural numbers // into two sets having GCD // of their sums greater than 1 using System;
class GFG{
// Function to create // and print the two sets static void createSets(long n)
{ // For n <= 2 such sets
// can never be formed
if (n <= 2)
{
Console.Write("-1");
return;
}
else
{
// Check if N is even or odd
// and store the element
// which divides the sum of N
// natural numbers accordingly
long x = (n % 2 == 0) ? (n / 2) :
((n + 1) / 2);
// First set
Console.Write(x + "\n");
// Print elements of second set
for(int i = 1; i <= n; i++)
{
if (i == x)
continue;
Console.Write(i + " ");
}
}
return;
} // Driver code public static void Main(String[] args)
{ long N = 7;
createSets(N);
} } // This code is contributed by sapnasingh4991 |
Javascript
<script> // Javascript program to split N natural numbers
// into two sets having GCD
// of their sums greater than 1
// Function to create
// and print the two sets
function createSets(n)
{
// For n <= 2 such sets
// can never be formed
if (n <= 2) {
document.write("-1");
return;
}
else {
// Check if N is even or odd
// and store the element
// which divides the sum of N
// natural numbers accordingly
let x = (n % 2 == 0) ? (n / 2) : ((n + 1) / 2);
// First set
document.write(x + "</br>");
// Print elements of second set
for (let i = 1; i <= n; i++)
{
if (i == x)
continue;
document.write(i + " ");
}
}
return;
}
// Driver code let N = 7;
createSets(N);
// This code is contributed by suresh07.
</script> |
Output:
4 1 2 3 5 6 7
Time Complexity: O(N)
Auxiliary Space Complexity: O(1)