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

  • Last Updated : 10 Jan, 2022

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: 2 4 6
Explanation: The triplet (2, 4, 6) is set of distinct integers such that 2 + 4 + 6 = 12 and 2 = GCD(4, 6).

Input: N = 5675
Output:1 2835 2839

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

Time Complexity: O(N2)
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)
 

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hrithik2108
@hrithik2108
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