Count distinct prime triplets up to N such that sum of two primes is equal to the third prime

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Given an integer N, the task is to count the number of distinct prime triplets (a, b, c) from the range [1, N] such that a < b < c ? N and a + b = c.
Note: Two prime tuples are distinct if at least one of the primes present in them are different.

Examples:

Input: N = 6
Output: 1
Explanation: Among numbers in the range [1, 6], the only prime triplet is (2, 3, 5) (Since 2 + 3 = 5).

Input: N = 10
Output: 2
Explanation: The distinct prime triplets satisfying the condition are (2, 3, 5), (2, 5, 7).

Approach: The problem can be solved based on the observation stated below:

Observation:

For every prime number p from 1 to N, it is a part of a triplet if and only if it can be represented as a sum of two prime numbers.
Since a prime number is an odd number, it must be equal to the sum of an even number and an odd number.
Hence the only even prime is 2. Therefore, for a prime number p to constitute a unique tuple (2, p-2, p), the number p – 2 must be a prime number.

Follow the steps below to solve the problem:

Initialize a variable, say count = 0, to store the number of prime triplets.
Iterate from 1 to N and for each number p, check if this number p and p – 2 are prime or not.
If they are prime, then increment count by 1.
Print the value of count.

Below is the implementation of the above approach:

C++

// C++ program for the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to check if a
// number is a prime or not
bool isPrime(int N)
{
    if (N <= 1)
        return false;
 
    for (int i = 2; i <= sqrt(N); i++) {
        if (N % i == 0)
            return false;
    }
 
    return true;
}
 
// Function to count the number
// of valid prime triplets
void countPrimeTuples(int N)
{
    // Stores the count
    // of prime triplets
    int count = 0;
 
    // Iterate from 2 to N and check for each
    // p, whether p & (p - 2) are prime or not
    for (int i = 2; i <= N; i++) {
 
        if (isPrime(i) && isPrime(i - 2))
            count++;
    }
 
    // Print the count obtained
    cout << count;
}
 
// Driver Code
int main()
{
    int N = 6;
    countPrimeTuples(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
   
  // Function to check if a
  // number is a prime or not
  static boolean isPrime(int N)
  {
      if (N <= 1)
          return false;
      for (int i = 2; i <= Math.sqrt(N); i++)
      {
          if (N % i == 0)
              return false;
      }
      return true;
  }
 
  // Function to count the number
  // of valid prime triplets
  static void countPrimeTuples(int N)
  {
     
      // Stores the count
      // of prime triplets
      int count = 0;
 
      // Iterate from 2 to N and check for each
      // p, whether p & (p - 2) are prime or not
      for (int i = 2; i <= N; i++) 
      {
          if (isPrime(i) && isPrime(i - 2))
              count++;
      }
 
      // Print the count obtained
      System.out.println(count);
  }
 
  // Driver Code
  public static void main (String[] args) 
  {  
    int N = 6;
    countPrimeTuples(N);
  }
}
 
// This code is contributed by Dharanendra L V.

Python3

# Python3 program for the above approach
import math 
 
# Function to check if a
# number is a prime or not
def isPrime(N) :
    if (N <= 1) :
        return False
    for i in range(2, int(math.sqrt(N) + 1)):
        if (N % i == 0) :
            return False
    return True
 
# Function to count the number
# of valid prime triplets
def countPrimeTuples(N) :
     
    # Stores the count
    # of prime triplets
    count = 0
 
    # Iterate from 2 to N and check for each
    # p, whether p & (p - 2) are prime or not
    for i in range(2, N + 1):
 
        if (isPrime(i) and isPrime(i - 2)) :
            count += 1
     
    # Print count obtained
    print(count)
 
# Driver Code
N = 6
countPrimeTuples(N)
 
# This code is contributed by susmitakundugoaldanga.

C#

// C# program for the above approach
using System;
public class GFG
{
     
  // Function to check if a
  // number is a prime or not
  static bool isPrime(int N)
  {
      if (N <= 1)
          return false;
      for (int i = 2; i <= Math.Sqrt(N); i++) 
      {
          if (N % i == 0)
              return false;
      }
      return true;
  }
 
  // Function to count the number
  // of valid prime triplets
  static void countPrimeTuples(int N)
  {
     
      // Stores the count
      // of prime triplets
      int count = 0;
 
      // Iterate from 2 to N and check for each
      // p, whether p & (p - 2) are prime or not
      for (int i = 2; i <= N; i++) 
      {
          if (isPrime(i) && isPrime(i - 2))
              count++;
      }
 
      // Print the count obtained
      Console.WriteLine(count);
  }
 
  // Driver Code
  static public void Main ()
  {
    int N = 6;
    countPrimeTuples(N);
  }
}
 
// This code is contributed by Dharanendra L V.

Javascript

<script>
 
// Javascript program for the above approach
 
// Function to check if a
// number is a prime or not
function isPrime(N)
{
    if (N <= 1)
        return false;
 
    for (let i = 2; i <= Math.sqrt(N); i++) {
        if (N % i == 0)
            return false;
    }
 
    return true;
}
 
// Function to count the number
// of valid prime triplets
function countPrimeTuples(N)
{
    // Stores the count
    // of prime triplets
    let count = 0;
 
    // Iterate from 2 to N and check for each
    // p, whether p & (p - 2) are prime or not
    for (let i = 2; i <= N; i++) {
 
        if (isPrime(i) && isPrime(i - 2))
            count++;
    }
 
    // Print the count obtained
    document.write(count);
}
 
// Driver Code
 
let N = 6;
countPrimeTuples(N);
 
// This code is contributed by _saurabh_jaiswal
 
</script>

Output:

Time Complexity: O(N^3/2)
Auxiliary Space: O(1)

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