C++ and Java Code for Sieve of Eratosthenes

The sieve of Eratosthenes is one of the most efficient ways to find all primes smaller than n when n is smaller than 10 million or so. Following is the algorithm to find all the prime numbers less than or equal to a given integer n by the Eratosthenes method:

When the algorithm terminates, all the numbers in the list that are not marked are prime.

• Firstly write all the numbers from 2,3,4…. n
• Now take the first prime number and mark all its multiples as visited.
• Now when you move forward take another number which is unvisited yet and then follow the same step-2 with that number.
• All numbers in the list left unmarked when the algorithm ends are referred to as prime numbers.

C++ Implementation

void SieveOfEratosthenes(int n)
{
    bool prime[n + 1];
    memset(prime, true, sizeof(prime));

    for (int p = 2; p * p <= n; p++)
    {
        if (prime[p] == true)
        {
            for (int i = p * p; i <= n; i += p)
                prime[i] = false;
        }
    }
    for (int p = 2; p <= n; p++)
        if (prime[p])
            cout << p << " ";
}

Java Implementation

class SieveOfEratosthenes {
    void sieveOfEratosthenes(int n)
    {
        boolean prime[] = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++){
            if (prime[p] == true)
            {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        for (int i = 2; i <= n; i++)
        {
            if (prime[i] == true)
                System.out.print(i + " ");
        }
    }
}

Time complexity of Sieve of Eratosthenes :
o(n * (log(log(n)))).

Space complexity: O(1)
Source - InterviewBit