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:Â
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.Â
Since a prime number is an odd number, it must be equal to the sum of an even number and an odd number.Â
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 approachÂ
#include <bits/stdc++.h>using namespace std;Â
// Function to check if a// number is a prime or notbool isPrime(int N){Â Â Â Â if (N <= 1)Â Â Â Â Â Â Â Â return false;Â
    for (int i = 2; i <= sqrt(N); i++) {        if (N % i == 0)            return false;    }Â
    return true;}Â
// Function to count the number// of valid prime tripletsvoid countPrimeTuples(int N){    // Stores the count    // of prime triplets    int count = 0;Â
    // 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++) {Â
        if (isPrime(i) && isPrime(i - 2))            count++;    }Â
    // Print the count obtained    cout << count;}Â
// Driver Codeint main(){Â Â Â Â int N = 6;Â Â Â Â countPrimeTuples(N);Â
    return 0;} |
Java
// Java program for the above approachimport 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;  }Â
  // Function to count the number  // of valid prime triplets  static void countPrimeTuples(int N)  {           // Stores the count      // of prime triplets      int count = 0;Â
      // 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++;      }Â
      // Print the count obtained      System.out.println(count);  }Â
  // Driver Code  public static void main (String[] args)   {     int N = 6;    countPrimeTuples(N);  }}Â
// This code is contributed by Dharanendra L V. |
Python3
# Python3 program for the above approachimport math Â
# Function to check if a# number is a prime or notdef isPrime(N) :    if (N <= 1) :        return False    for i in range(2, int(math.sqrt(N) + 1)):        if (N % i == 0) :            return False    return TrueÂ
# Function to count the number# of valid prime tripletsdef countPrimeTuples(N) :         # Stores the count    # of prime triplets    count = 0Â
    # 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):Â
        if (isPrime(i) and isPrime(i - 2)) :            count += 1         # Print count obtained    print(count)Â
# Driver CodeN = 6countPrimeTuples(N)Â
# This code is contributed by susmitakundugoaldanga. |
C#
// C# program for the above approachusing 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;  }Â
  // Function to count the number  // of valid prime triplets  static void countPrimeTuples(int N)  {           // Stores the count      // of prime triplets      int count = 0;Â
      // 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++;      }Â
      // Print the count obtained      Console.WriteLine(count);  }Â
  // Driver Code  static public void Main ()  {    int N = 6;    countPrimeTuples(N);  }}Â
// This code is contributed by Dharanendra L V. |
Javascript
<script>Â
// Javascript program for the above approachÂ
// Function to check if a// number is a prime or notfunction isPrime(N){Â Â Â Â if (N <= 1)Â Â Â Â Â Â Â Â return false;Â
    for (let i = 2; i <= Math.sqrt(N); i++) {        if (N % i == 0)            return false;    }Â
    return true;}Â
// Function to count the number// of valid prime tripletsfunction countPrimeTuples(N){    // Stores the count    // of prime triplets    let count = 0;Â
    // 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++) {Â
        if (isPrime(i) && isPrime(i - 2))            count++;    }Â
    // Print the count obtained    document.write(count);}Â
// Driver CodeÂ
let N = 6;countPrimeTuples(N);Â
// This code is contributed by _saurabh_jaiswalÂ
</script> |
Output:Â
1
Â
Time Complexity: O(N3/2)
Auxiliary Space: O(1)Â
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