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Count prime pairs whose difference is also a Prime Number
  • Last Updated : 23 Apr, 2021

Given an integer N, the task is to count the number of pairs of prime numbers in the range [1, N] such that the difference between elements of each pair is also a prime number.

Examples:

Input: N = 5 
Output:
Explanations: 
Pair of prime numbers in the range [1, 5] whose difference between elements is also a prime number are: 
(2, 5) = 3 (Prime number) 
(3, 5) = 2 (Prime number) 
Therefore, the count of pairs of the prime numbers whose difference is also a prime number is 2. 

Input: N = 11 
Output: 4

 

Naive Approach: The simplest approach to solve this problem is to generate all possible pairs of the elements in the range [1, N] and for each pair, check if both the elements and the difference between both the elements of the pair is a prime number or not. If found to be true then increment the count. Finally, print the count.



Time Complexity: O(N2 * √N) 
Auxiliary Space: O(1)

Efficient Approach: To optimize the above approach, the idea is based on the following observations:

Odd number – Even Number = Odd Number 
Odd number – Odd number = Even number 
2 is the only even prime number. 
Therefore, the problem reduces to check only for those pairs of prime numbers whose difference between the elements of the pair is equal to 2. 
 

Follow the steps below to solve the problem:

  • Initialize a variable, say cntPairs to store the count of pairs of prime numbers such that the difference between elements of each pair is also a prime number.
  • Initialize an array, say sieve[] to check if a number in the range [1, N] is a prime number or not.
  • Find all the prime numbers in the range [1, N] using Sieve of Eratosthenes.
  • Iterate over the range [2, N], and for each element in the given range, check if the sieve[i] and sieve[i – 2] are true or not. If found to be true then increment the value of cntPairs by 2.
  • Finally, print the value of cntPairs.

Below is the implementation of the above approach:

C++




// C++ program to implement
// the above approach
 
#include <bits/stdc++.h>
using namespace std;
 
// Function to find all prime
// numbers in the range [1, N]
vector<bool> SieveOfEratosthenes(
    int N)
{
   
    // isPrime[i]: Stores if i is
    // a prime number or not
    vector<bool> isPrime(N, true);
   
    isPrime[0] = false;
    isPrime[1] = false;
   
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for (int p = 2; p * p <= N; p++) {
   
        // If P is a prime number
        if (isPrime[p]) {
   
            // Set all multiple of P
            // as non-prime
            for (int i = p * p; i <= N;
                 i += p) {
   
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
 
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
int cntPairsdiffOfPrimeisPrime(int N)
{
     
    // Function to count pairs of
    // prime numbers whose difference  
    // is also a prime number
    int cntPairs = 0;
     
     
    // isPrime[i]: Stores if i is
    // a prime number or not
    vector<bool> isPrime
          = SieveOfEratosthenes(N);
     
    // Iterate over the range [2, N]
    for (int i = 2; i <= N; i++) {
         
         
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] &&
            isPrime[i - 2]) {
               
               
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
 
// Driver Code
int main()
{
    int N = 5;
    cout << cntPairsdiffOfPrimeisPrime(N);
    return 0;
}

Java




// Java program to implement
// the above approach
import java.util.*;
 
class GFG{
     
// Function to find all prime
// numbers in the range [1, N]
public static boolean[] SieveOfEratosthenes(int N)
{
     
    // isPrime[i]: Stores if i is
    // a prime number or not
    boolean[] isPrime = new boolean[N + 1];
    Arrays.fill(isPrime, true);
 
    isPrime[0] = false;
    isPrime[1] = false;
 
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for(int p = 2; p * p <= N; p++)
    {
         
        // If P is a prime number
        if (isPrime[p])
        {
             
            // Set all multiple of P
            // as non-prime
            for(int i = p * p; i <= N; i += p)
            {
                 
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
 
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
public static int cntPairsdiffOfPrimeisPrime(int N)
{
     
    // Function to count pairs of
    // prime numbers whose difference
    // is also a prime number
    int cntPairs = 0;
 
    // isPrime[i]: Stores if i is
    // a prime number or not
    boolean[] isPrime = SieveOfEratosthenes(N);
 
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] && isPrime[i - 2])
        {
             
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
 
// Driver Code
public static void main(String args[])
{
    int N = 5;
     
    System.out.println(cntPairsdiffOfPrimeisPrime(N));
}
}
 
// This code is contributed by hemanth gadarla

Python3




# Python3 program to implement
# the above approach
from math import sqrt
 
# Function to find all prime
# numbers in the range [1, N]
def SieveOfEratosthenes(N):
     
    # isPrime[i]: Stores if i is
    # a prime number or not
    isPrime = [True for i in range(N + 1)]
   
    isPrime[0] = False
    isPrime[1] = False
   
    # Calculate all prime numbers up to
    # Max using Sieve of Eratosthenes
    for p in range(2, int(sqrt(N)) + 1, 1):
         
        # If P is a prime number
        if (isPrime[p]):
             
            # Set all multiple of P
            # as non-prime
            for i in range(p * p, N + 1, p):
                 
                # Update isPrime
                isPrime[i] = False
                 
    return isPrime
 
# Function to count pairs of
# prime numbers in the range [1, N]
# whose difference is prime
def cntPairsdiffOfPrimeisPrime(N):
     
    # Function to count pairs of
    # prime numbers whose difference  
    # is also a prime number
    cntPairs = 0
     
    # isPrime[i]: Stores if i is
    # a prime number or not
    isPrime = SieveOfEratosthenes(N)
     
    # Iterate over the range [2, N]
    for i in range(2, N + 1, 1):
         
        # If i and i - 2 is
        # a prime number
        if (isPrime[i] and isPrime[i - 2]):
             
            # Update cntPairs
            cntPairs += 2
 
    return cntPairs
 
# Driver Code
if __name__ == '__main__':
     
    N = 5
     
    print(cntPairsdiffOfPrimeisPrime(N))
 
# This code is contributed by ipg2016107

C#




// C# program to implement
// the above approach 
using System;
  
class GFG{
     
// Function to find all prime
// numbers in the range [1, N]
public static bool[] SieveOfEratosthenes(int N)
{
     
    // isPrime[i]: Stores if i is
    // a prime number or not
    bool[] isPrime = new bool[N + 1];
    for(int i = 0; i < N + 1; i++)
    {
        isPrime[i] = true;
    }
  
    isPrime[0] = false;
    isPrime[1] = false;
  
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for(int p = 2; p * p <= N; p++)
    {
         
        // If P is a prime number
        if (isPrime[p])
        {
             
            // Set all multiple of P
            // as non-prime
            for(int i = p * p; i <= N; i += p)
            {
                 
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
  
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
public static int cntPairsdiffOfPrimeisPrime(int N)
{
     
    // Function to count pairs of
    // prime numbers whose difference
    // is also a prime number
    int cntPairs = 0;
  
    // isPrime[i]: Stores if i is
    // a prime number or not
    bool[] isPrime = SieveOfEratosthenes(N);
  
    // Iterate over the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] && isPrime[i - 2])
        {
             
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
  
// Driver Code
public static void Main()
{
    int N = 5;
      
    Console.WriteLine(cntPairsdiffOfPrimeisPrime(N));
}
}
 
// This code is contributed by susmitakundugoaldanga

Javascript




<script>
 
// JavaScript program to implement
// the above approach
 
// Function to find all prime
// numbers in the range [1, N]
function SieveOfEratosthenes(N)
{
      
    // isPrime[i]: Stores if i is
    // a prime number or not
    let isPrime = [];
    for(let i = 0; i < N + 1; i++)
    {
        isPrime[i] = true;
    }
  
    isPrime[0] = false;
    isPrime[1] = false;
  
    // Calculate all prime numbers up to
    // Max using Sieve of Eratosthenes
    for(let p = 2; p * p <= N; p++)
    {
          
        // If P is a prime number
        if (isPrime[p])
        {
              
            // Set all multiple of P
            // as non-prime
            for(let i = p * p; i <= N; i += p)
            {
                  
                // Update isPrime
                isPrime[i] = false;
            }
        }
    }
    return isPrime;
}
  
// Function to count pairs of
// prime numbers in the range [1, N]
// whose difference is prime
function cntPairsdiffOfPrimeisPrime(N)
{
      
    // Function to count pairs of
    // prime numbers whose difference
    // is also a prime number
    let cntPairs = 0;
  
    // isPrime[i]: Stores if i is
    // a prime number or not
    let isPrime = SieveOfEratosthenes(N);
  
    // Iterate over the range [2, N]
    for(let i = 2; i <= N; i++)
    {
          
        // If i and i - 2 is
        // a prime number
        if (isPrime[i] && isPrime[i - 2])
        {
              
            // Update cntPairs
            cntPairs += 2;
        }
    }
    return cntPairs;
}
 
// Driver Code
 
    let N = 5;
      
    document.write(cntPairsdiffOfPrimeisPrime(N));
      
</script>
Output: 
2

 

Time Complexity: O(N * log(log(N))) 
Auxiliary Space: O(N)

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