Find a triplet (X, Y, Z) with given sum as N and GCD of two numbers is the third number

Given a positive integer N, the task is to find a triple of three distinct positive integers (X, Y, Z) such that X + Y + Z = N and X = GCD (Y, Z).

Example:

Input: N = 12
Output:1 2 9
Explanation: The triplet (1, 2, 9) is set of distinct integers such that 1 + 2 + 9 = 12 and 1 = GCD(2, 9).

Input: N = 5675
Output:1 3 5671

Naive Approach: The basic idea is to iterate to find all possible triplets of (X, Y, Z) with sum N and for each triplet, check if GCD(Y, Z) = X.

Code-

C++

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

// Function to find a triplet (X, Y, Z)
// of distinct integers with their sum
// as N and GCD(Y, Z) = X
void printTriplet(int N)
{
    for(int i=1;i<N;i++){
        for(int j=i+1;j<N;j++){
            for(int k=j+1;k<N;k++){
                if(i+j+k==N && __gcd(j,k)==i){
                    cout<<i<<" "<<j<<" "<<k<<endl;
                    return;
                }
            }
        }
    }
}

// Driver Code
int main()
{
    int N = 5875;
    printTriplet(N);

    return 0;
}

Java

import java.util.*;

class GFG
{

  // Function to find a triplet (X, Y, Z)
  // of distinct integers with their sum
  // as N and GCD(Y, Z) = X
  static void printTriplet(int N)
  {
    for (int i = 1; i < N; i++) {
      for (int j = i + 1; j < N; j++) {
        for (int k = j + 1; k < N; k++) {
          if (i + j + k == N && gcd(j, k) == i) {
            System.out.println(i + " " + j + " "
                               + k);
            return;
          }
        }
      }
    }
  }

  // Function to calculate GCD of two numbers
  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 = 5875;
    printTriplet(N);
  }
}

Python3

import math

# Function to find a triplet (X, Y, Z)
# of distinct integers with their sum
# as N and GCD(Y, Z) = X


def printTriplet(N):
    for i in range(1, N):
        for j in range(i+1, N):
            for k in range(j+1, N):
                if i+j+k == N and math.gcd(j, k) == i:
                    print(i, j, k)
                    return


# Driver Code
if __name__ == "__main__":
    N = 5875
    printTriplet(N)

C#

using System;

class MainClass {
    // Function to find a triplet (X, Y, Z)
    // of distinct integers with their sum
    // as N and GCD(Y, Z) = X
    static void PrintTriplet(int N)
    {
        for (int i = 1; i < N; i++) {
            for (int j = i + 1; j < N; j++) {
                for (int k = j + 1; k < N; k++) {
                    if (i + j + k == N && gcd(j, k) == i) {
                        Console.WriteLine(i + " " + j + " "
                                          + k);
                        return;
                    }
                }
            }
        }
    }

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

    // Driver Code
    public static void Main(string[] args)
    {
        int N = 5875;
        PrintTriplet(N);
    }
}

Javascript

// Function to find a triplet (X, Y, Z)
// of distinct integers with their sum
// as N and GCD(Y, Z) = X

function printTriplet(N)
{
  for (let i = 1; i < N; i++)
  {
    for (let j = i + 1; j < N; j++)
    {
      for (let k = j + 1; k < N; k++)
      {
        if (i + j + k == N && gcd(j, k) == i)
        {
          console.log(i, j, k);
          return;
        }
      }
    }
  }
}

// Function to find the greatest common divisor (GCD) of two numbers
function gcd(a, b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}

// Driver Code
let N = 5875;
printTriplet(N);

Output

1 5 5869

Time Complexity: O(N³*logN),logN for finding GCD, and N³ for three “for” loops
Auxiliary space: O(1)

Efficient Approach: The above approach can be further optimized using the observation that for any given N, there are the following three cases:

• Case 1: If N is even then, a valid triplet is (1, N/2, N/2 -1).
• Case 2: If N is odd and (N/2) is even then, a valid triplet is (1, N/2 + 1, N/2 -1).
• Case 3: If N is odd and (N/2) is also odd then, a valid triplet is (1, N/2 – 2, N/2 + 2).

Hence, for any given N, identify the case and print its respective triplet.
Below is the implementation of the approach:

C++

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

// Function to find a triplet (X, Y, Z)
// of distinct integers with their sum
// as N and GCD(Y, Z) = X
int printTriplet(int N)
{
    // Case 1 where N is even
    if (N % 2 == 0) {
        cout << 1 << " " << (N / 2)
<< " " << (N / 2) - 1;
    }
    else {

        // Case 2 where N is Odd
        // and N/2 is even
        if ((N / 2) % 2 == 0) {
            cout << 1 << " "
 << (N / 2) - 1 << " "
                 << (N / 2) + 1;
        }

        // Case 3 where N is Odd
        // and N/2 is also odd
        else {
            cout << 1 << " "
 << (N / 2) - 2 << " "
                 << (N / 2) + 2;
        }
    }
}

// Driver Code
int main()
{
    int N = 5875;
    printTriplet(N);

    return 0;
}

Java

// Java program of the above approach
import java.util.*;
class GFG
{

  // Function to find a triplet (X, Y, Z)
  // of distinct integers with their sum
  // as N and GCD(Y, Z) = X
  static void printTriplet(int N)
  {

    // Case 1 where N is even
    if (N % 2 == 0) {
      System.out.print(1 + " " + (N / 2) +
                       " " + ((N / 2) - 1));
    } else {

      // Case 2 where N is Odd
      // and N/2 is even
      if ((N / 2) % 2 == 0) {
        System.out.print(1 + " " + ((N / 2) - 1) +
                         " " + ((N / 2) + 1));
      }

      // Case 3 where N is Odd
      // and N/2 is also odd
      else {
        System.out.print(1 + " " + ((N / 2) - 2) +
                         " " + ((N / 2) + 2));
      }
    }
  }

  // Driver Code
  public static void main(String[] args) {
    int N = 5875;
    printTriplet(N);

  }
}

// This code is contributed by 29AjayKumar

Python3

# python3 program of the above approach

# Function to find a triplet (X, Y, Z)
# of distinct integers with their sum
# as N and GCD(Y, Z) = X
def printTriplet(N):

    # Case 1 where N is even
    if (N % 2 == 0):
        print(f"{1} {(N / 2)} {(N / 2) - 1}")

    else:

        # Case 2 where N is Odd
        # and N/2 is even
        if ((N // 2) % 2 == 0):
            print(f"{1} {(N // 2) - 1} {(N // 2) + 1}")

        # Case 3 where N is Odd
        # and N/2 is also odd
        else:
            print(f"{1} {(N // 2) - 2} {(N // 2) + 2}")

# Driver Code
if __name__ == "__main__":

    N = 5875
    printTriplet(N)

# This code is contributed by rakeshsahni

C#

// C# program of the above approach
using System;

class GFG{

// Function to find a triplet (X, Y, Z)
// of distinct integers with their sum
// as N and GCD(Y, Z) = X
static void printTriplet(int N)
{

    // Case 1 where N is even
    if (N % 2 == 0)
    {
        Console.Write(1 + " " + (N / 2) + " " +
                               ((N / 2) - 1));
    }
    else
    {

        // Case 2 where N is Odd
        // and N/2 is even
        if ((N / 2) % 2 == 0)
        {
            Console.Write(1 + " " + ((N / 2) - 1) + " " +
                                    ((N / 2) + 1));
        }

        // Case 3 where N is Odd
        // and N/2 is also odd
        else
        {
            Console.Write(1 + " " + ((N / 2) - 2) + " " +
                                    ((N / 2) + 2));
        }
    }
}

// Driver Code
public static void Main()
{
    int N = 5875;

    printTriplet(N);
}
}

// This code is contributed by ukasp

Javascript

<script>
       // JavaScript code for the above approach

       // Function to find a triplet (X, Y, Z)
       // of distinct integers with their sum
       // as N and GCD(Y, Z) = X
       function printTriplet(N)
       {

           // Case 1 where N is even
           if (N % 2 == 0) {
               document.write(1 + " " + Math.floor(N / 2)
                   + " " + (Math.floor(N / 2) - 1));
           }
           else {

               // Case 2 where N is Odd
               // and N/2 is even
               if ((N / 2) % 2 == 0) {
                   document.write(1 + " "
                       + (Math.floor(N / 2) - 1) + " "
                       + (Math.floor(N / 2) + 1));
               }

               // Case 3 where N is Odd
               // and N/2 is also odd
               else {
                   document.write(1 + " "
                       + (Math.floor(N / 2) - 2) + " "
                       + (Math.floor(N / 2) + 2));
               }
           }
       }

       // Driver Code
       let N = 5875;
       printTriplet(N);

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

Output

1 2935 2939

Time Complexity: O(1)
Auxiliary space: O(1)