Count distinct prime triplets up to N such that sum of two primes is equal to the third prime
Last Updated :
15 Nov, 2021
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:
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
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 ;
}
void countPrimeTuples( int N)
{
int count = 0;
for ( int i = 2; i <= N; i++) {
if (isPrime(i) && isPrime(i - 2))
count++;
}
cout << count;
}
int main()
{
int N = 6;
countPrimeTuples(N);
return 0;
}
|
Java
import java.io.*;
class GFG
{
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 ;
}
static void countPrimeTuples( int N)
{
int count = 0 ;
for ( int i = 2 ; i <= N; i++)
{
if (isPrime(i) && isPrime(i - 2 ))
count++;
}
System.out.println(count);
}
public static void main (String[] args)
{
int N = 6 ;
countPrimeTuples(N);
}
}
|
Python3
import math
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
def countPrimeTuples(N) :
count = 0
for i in range ( 2 , N + 1 ):
if (isPrime(i) and isPrime(i - 2 )) :
count + = 1
print (count)
N = 6
countPrimeTuples(N)
|
C#
using System;
public class GFG
{
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 ;
}
static void countPrimeTuples( int N)
{
int count = 0;
for ( int i = 2; i <= N; i++)
{
if (isPrime(i) && isPrime(i - 2))
count++;
}
Console.WriteLine(count);
}
static public void Main ()
{
int N = 6;
countPrimeTuples(N);
}
}
|
Javascript
<script>
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 countPrimeTuples(N)
{
let count = 0;
for (let i = 2; i <= N; i++) {
if (isPrime(i) && isPrime(i - 2))
count++;
}
document.write(count);
}
let N = 6;
countPrimeTuples(N);
</script>
|
Time Complexity: O(N3/2)
Auxiliary Space: O(1)
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