Split given number into two distinct numbers having GCD 1

Given an Integer as N, the task is to convert N into two numbers a and b, such that a < b and GCD(a, b) is equal to 1.

Examples:

Input: N = 12
Output: 5 7

Input: N = 9
Output: 4 5

Approach: The idea is to start iterating from the back of the loop starting from the number itself. Follow the steps given below to solve the problem.

• If N is less than equal to 2, the answer doesn’t exist.
• Otherwise iterate over the range [N/2, 0) using the variable i and perform the following tasks:
  • If the gcd of i and N-i is 1 then print this as the answer and return.

Below is the implementation of the above approach:

C++

/// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;

// Function to split N into two numbers
void find(int n)
{

    // Base Case
    if (n <= 2) {
        cout << "-1";
    }
    else {

        // Since a < b
        // start from n/2
        for (int i = n / 2; i > 0; i--) {

            // Sum of a and b is n
            int a = i, b = n - i;

            // Check the gcd
            if (__gcd(a, b) == 1 && a < b) {
                cout << a <<", " << b;
                return;
            }
        }
    }
}

// Driver Code
int main()
{

    int N = 9;

    find(N);

    return 0;
}

Java

/// Java program for the above approach
class GFG {

  // Function to split N into two numbers
  public static void find(int n) {

    // Base Case
    if (n <= 2) {
      System.out.println("-1");
    } else {

      // Since a < b
      // start from n/2
      for (int i = n / 2; i > 0; i--) {

        // Sum of a and b is n
        int a = i, b = n - i;

        // Check the gcd
        if (__gcd(a, b) == 1 && a < b) {
          System.out.println(a + ", " + b);
          return;
        }
      }
    }
  }

  static int __gcd(int a, int b) {
    if (b == 0)
      return a;
    return __gcd(b, a % b);
  }

  // Driver Code
  public static void main(String args[]) {
    int N = 9;
    find(N);
  }
}

// This code is contributed by saurabh_jaiswal.

Python3

# Python 3 program for the above approach
import math

# Function to split N into two numbers
def find(n):

    # Base Case
    if (n <= 2):
        print("-1")

    else:

        # Since a < b
        # start from n/2
        for i in range(n // 2, -1, -1):

            # Sum of a and b is n
            a = i
            b = n - i

            # Check the gcd
            if (math.gcd(a, b) == 1 and a < b):
                print(a,",", b)
                return

# Driver Code
if __name__ == "__main__":

    N = 9
    find(N)

    #vThis code is contributed by ukasp.

C#

/// C# program for the above approach
using System;
public class GFG {

  // Function to split N into two numbers
  public static void find(int n) {

    // Base Case
    if (n <= 2) {
      Console.WriteLine("-1");
    } else {

      // Since a < b
      // start from n/2
      for (int i = n / 2; i > 0; i--) {

        // Sum of a and b is n
        int a = i, b = n - i;

        // Check the gcd
        if (__gcd(a, b) == 1 && a < b) {
          Console.WriteLine(a + ", " + b);
          return;
        }
      }
    }
  }

  static int __gcd(int a, int b) {
    if (b == 0)
      return a;
    return __gcd(b, a % b);
  }

  // Driver Code
  public static void Main(String []args) {
    int N = 9;
    find(N);
  }
}

// This code is contributed by 29AjayKumar

Javascript

<script>
       // JavaScript code for the above approach

       // Recursive function to return gcd of a and b
       function __gcd(a, b) {
           // Everything divides 0
           if (a == 0)
               return b;
           if (b == 0)
               return a;

           // base case
           if (a == b)
               return a;

           // a is greater
           if (a > b)
               return __gcd(a - b, b);
           return __gcd(a, b - a);
       }
       // Function to split N into two numbers
       function find(n) {

           // Base Case
           if (n <= 2) {
               document.write("-1");
           }
           else {

               // Since a < b
               // start from n/2
               for (let i = Math.floor(n / 2); i > 0; i--) {

                   // Sum of a and b is n
                   let a = i, b = n - i;

                   // Check the gcd
                   if (__gcd(a, b) == 1 && a < b) {
                       document.write(a + ", " + b);
                       return;
                   }
               }
           }
       }

       // Driver Code
       let N = 9;
       find(N);

 // This code is contributed by Potta Lokesh
   </script>

Output

4, 5

Time complexity: O(N log(N))
Auxiliary space: O(1)