Count of primes below N which can be expressed as the sum of two primes
Given an integer N, the task is to find the count of all the primes below N which can be expressed as the sum of two primes.
Examples:
Input: N = 6
Output: 1
5 is the only such prime below 6.
2 + 3 = 5.
Input: N = 11
Output: 2
Approach: Create an array prime[] where prime[i] will store whether i is prime or not using Sieve of Eratosthenes. Now for every prime from the range [1, N – 1], check whether it can be expressed as the sum of two primes using the approach discussed here.
Below is the implementation of the above approach:
C++
// C++ implementation of the approach#include <bits/stdc++.h>using namespace std;const int MAX = 100005;bool prime[MAX];// Function for Sieve of Eratosthenesvoid SieveOfEratosthenes(){ memset(prime, true, sizeof(prime)); // false here indicates // that it is not prime prime[0] = false; prime[1] = false; for (int p = 2; p * p <= MAX; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p]) { // Update all multiples of p, // set them to non-prime for (int i = p * 2; i <= MAX; i += p) prime[i] = false; } }}// Function to return the count of primes// less than or equal to n which can be// expressed as the sum of two primesint countPrimes(int n){ SieveOfEratosthenes(); // To store the required count int cnt = 0; for (int i = 2; i < n; i++) { // If the integer is prime and it // can be expressed as the sum of // 2 and a prime number if (prime[i] && prime[i - 2]) cnt++; } return cnt;}// Driver codeint main(){ int n = 11; cout << countPrimes(n); return 0;} |
Java
// Java implementation of the approachclass GFG{static int MAX = 100005;static boolean []prime = new boolean[MAX];// Function for Sieve of Eratosthenesstatic void SieveOfEratosthenes(){ for (int i = 0; i < MAX; i++) prime[i] = true; // false here indicates // that it is not prime prime[0] = false; prime[1] = false; for (int p = 2; p * p < MAX; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p]) { // Update all multiples of p, // set them to non-prime for (int i = p * 2; i < MAX; i += p) prime[i] = false; } }}// Function to return the count of primes// less than or equal to n which can be// expressed as the sum of two primesstatic int countPrimes(int n){ SieveOfEratosthenes(); // To store the required count int cnt = 0; for (int i = 2; i < n; i++) { // If the integer is prime and it // can be expressed as the sum of // 2 and a prime number if (prime[i] && prime[i - 2]) cnt++; } return cnt;}// Driver codepublic static void main(String[] args){ int n = 11; System.out.print(countPrimes(n));}}// This code is contributed by 29AjayKumar |
Python3
# Python3 implementation of the approachMAX = 100005prime = [True for i in range(MAX)]# Function for Sieve of Eratosthenesdef SieveOfEratosthenes(): # False here indicates # that it is not prime prime[0] = False prime[1] = False for p in range(MAX): if(p * p > MAX): break # If prime[p] is not changed, # then it is a prime if (prime[p]): # Update all multiples of p, # set them to non-prime for i in range(2 * p, MAX, p): prime[i] = False# Function to return the count of primes# less than or equal to n which can be# expressed as the sum of two primesdef countPrimes(n): SieveOfEratosthenes() # To store the required count cnt = 0 for i in range(2, n): # If the integer is prime and it # can be expressed as the sum of # 2 and a prime number if (prime[i] and prime[i - 2]): cnt += 1 return cnt# Driver coden = 11print(countPrimes(n))# This code is contributed by Mohit Kumar |
C#
// C# implementation of the approachusing System;class GFG{static int MAX = 100005;static bool []prime = new bool[MAX];// Function for Sieve of Eratosthenesstatic void SieveOfEratosthenes(){ for (int i = 0; i < MAX; i++) prime[i] = true; // false here indicates // that it is not prime prime[0] = false; prime[1] = false; for (int p = 2; p * p < MAX; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p]) { // Update all multiples of p, // set them to non-prime for (int i = p * 2; i < MAX; i += p) prime[i] = false; } }}// Function to return the count of primes// less than or equal to n which can be// expressed as the sum of two primesstatic int countPrimes(int n){ SieveOfEratosthenes(); // To store the required count int cnt = 0; for (int i = 2; i < n; i++) { // If the integer is prime and it // can be expressed as the sum of // 2 and a prime number if (prime[i] && prime[i - 2]) cnt++; } return cnt;}// Driver codepublic static void Main(String[] args){ int n = 11; Console.Write(countPrimes(n));}}// This code is contributed by Rajput-Ji |
Javascript
<script>// javascript implementation of the approach var MAX = 100005;var prime = new Array(MAX).fill(false); // Function for Sieve of Eratosthenes function SieveOfEratosthenes() { for (i = 0; i < MAX; i++) prime[i] = true; // false here indicates // that it is not prime prime[0] = false; prime[1] = false; for (p = 2; p * p < MAX; p++) { // If prime[p] is not changed, // then it is a prime if (prime[p]) { // Update all multiples of p, // set them to non-prime for (i = p * 2; i < MAX; i += p) prime[i] = false; } } } // Function to return the count of primes // less than or equal to n which can be // expressed as the sum of two primes function countPrimes(n) { SieveOfEratosthenes(); // To store the required count var cnt = 0; for (i = 2; i < n; i++) { // If the integer is prime and it // can be expressed as the sum of // 2 and a prime number if (prime[i] && prime[i - 2]) cnt++; } return cnt; } // Driver code var n = 11; document.write(countPrimes(n));// This code is contributed by todaysgaurav</script> |
Output:
2
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