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

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