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 Eratosthenes void 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 primes 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 code int main() { int n = 11; cout << countPrimes(n); return 0; } |
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Java
// Java implementation of the approach class GFG { static int MAX = 100005; static boolean []prime = new boolean[MAX]; // Function for Sieve of Eratosthenes static 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 primes static 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 code public static void main(String[] args) { int n = 11; System.out.print(countPrimes(n)); } } // This code is contributed by 29AjayKumar |
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Python3
# Python3 implementation of the approach MAX = 100005prime = [True for i in range(MAX)] # Function for Sieve of Eratosthenes def 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 primes def 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 code n = 11 print(countPrimes(n)) # This code is contributed by Mohit Kumar |
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C#
// C# implementation of the approach using System; class GFG { static int MAX = 100005; static bool []prime = new bool[MAX]; // Function for Sieve of Eratosthenes static 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 primes static 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 code public static void Main(String[] args) { int n = 11; Console.Write(countPrimes(n)); } } // This code is contributed by Rajput-Ji |
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Output:
2
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