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: 2
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 primeint 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 Codeint main(){ int N = 5; cout << cntPairsdiffOfPrimeisPrime(N); return 0;} |
Java
// Java program to implement// the above approachimport 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 primepublic 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 Codepublic 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 approachfrom 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 primedef 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 Codeif __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 primepublic 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 Codepublic 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 primefunction 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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