Sieve of Eratosthenes in 0(n) time complexity

The classical Sieve of Eratosthenes algorithm takes O(N log (log N)) time to find all prime numbers less than N. In this article, a modified Sieve is discussed that works in O(N) time.

Example :

Given a number N, print all prime 
numbers smaller than N

Input :  int N = 15
Output : 2 3 5 7 11 13

Input : int N = 20
Output : 2 3 5 7 11 13 17 19

Manipulated Sieve of Eratosthenes algorithm works as following:

For every number i where i varies from 2 to N-1:
    Check if the number is prime. If the number
    is prime, store it in prime array.

For every prime numbers j less than or equal to the smallest  
prime factor p of i:
    Mark all numbers j*p as non_prime.
    Mark smallest prime factor of j*p as j

Below is implementation of above idea.

C++

// C++ program to generate all prime numbers 
// less than N in O(N) 
#include<bits/stdc++.h> 
using namespace std; 
const long long MAX_SIZE = 1000001; 
  
// isPrime[] : isPrime[i] is true if number is prime  
// prime[] : stores all prime number less than N 
// SPF[] that store smallest prime factor of number 
// [for Exp : smallest prime factor of '8' and '16' 
// is '2' so we put SPF[8] = 2 , SPF[16] = 2 ] 
vector<long long >isprime(MAX_SIZE , true); 
vector<long long >prime; 
vector<long long >SPF(MAX_SIZE); 
  
// function generate all prime number less then N in O(n) 
void manipulated_seive(int N) 
{ 
    // 0 and 1 are not prime 
    isprime[0] = isprime[1] = false ; 
  
    // Fill rest of the entries 
    for (long long int i=2; i<N ; i++) 
    { 
        // If isPrime[i] == True then i is 
        // prime number 
        if (isprime[i]) 
        { 
            // put i into prime[] vector 
            prime.push_back(i); 
  
            // A prime number is its own smallest 
            // prime factor 
            SPF[i] = i; 
        } 
  
        // Remove all multiples of  i*prime[j] which are 
        // not prime by making isPrime[i*prime[j]] = false 
        // and put smallest prime factor of i*Prime[j] as prime[j] 
        // [ for exp :let  i = 5 , j = 0 , prime[j] = 2 [ i*prime[j] = 10 ] 
        // so smallest prime factor of '10' is '2' that is prime[j] ] 
        // this loop run only one time for number which are not prime 
        for (long long int j=0; 
             j < (int)prime.size() && 
             i*prime[j] < N && prime[j] <= SPF[i]; 
             j++) 
        { 
            isprime[i*prime[j]]=false; 
  
            // put smallest prime factor of i*prime[j] 
            SPF[i*prime[j]] = prime[j] ; 
        } 
    } 
} 
  
// driver  program to test above function 
int main() 
{ 
    int N = 13 ; // Must be less than MAX_SIZE 
  
    manipulated_seive(N); 
  
    // pint all prime number less then N 
    for (int i=0; i<prime.size() && prime[i] <= N ; i++) 
        cout << prime[i] << " "; 
  
    return 0; 
} 

Java

// Java program to generate all prime numbers 
// less than N in O(N) 
  
  
import java.util.Vector; 
  
class Test 
{ 
    static final int MAX_SIZE = 1000001; 
    // isPrime[] : isPrime[i] is true if number is prime  
    // prime[] : stores all prime number less than N 
    // SPF[] that store smallest prime factor of number 
    // [for Exp : smallest prime factor of '8' and '16' 
    // is '2' so we put SPF[8] = 2 , SPF[16] = 2 ] 
    static Vector<Boolean>isprime = new Vector<>(MAX_SIZE); 
    static Vector<Integer>prime = new Vector<>(); 
    static Vector<Integer>SPF = new Vector<>(MAX_SIZE); 
       
    // method generate all prime number less then N in O(n) 
    static void manipulated_seive(int N) 
    { 
        // 0 and 1 are not prime 
        isprime.set(0, false); 
        isprime.set(1, false); 
          
        // Fill rest of the entries 
        for (int i=2; i<N ; i++) 
        { 
            // If isPrime[i] == True then i is 
            // prime number 
            if (isprime.get(i)) 
            { 
                // put i into prime[] vector 
                prime.add(i); 
       
                // A prime number is its own smallest 
                // prime factor 
                SPF.set(i,i); 
            } 
       
            // Remove all multiples of  i*prime[j] which are 
            // not prime by making isPrime[i*prime[j]] = false 
            // and put smallest prime factor of i*Prime[j] as prime[j] 
            // [ for exp :let  i = 5 , j = 0 , prime[j] = 2 [ i*prime[j] = 10 ] 
            // so smallest prime factor of '10' is '2' that is prime[j] ] 
            // this loop run only one time for number which are not prime 
            for (int j=0; 
                 j < prime.size() && 
                 i*prime.get(j) < N && prime.get(j) <= SPF.get(i); 
                 j++) 
            { 
                isprime.set(i*prime.get(j),false); 
       
                // put smallest prime factor of i*prime[j] 
                SPF.set(i*prime.get(j),prime.get(j)) ; 
            } 
        } 
    } 
      
    // Driver method 
    public static void main(String args[])  
    { 
        int N = 13 ; // Must be less than MAX_SIZE 
          
        // initializing isprime and spf 
        for (int i = 0; i < MAX_SIZE; i++){ 
            isprime.add(true); 
            SPF.add(2); 
        } 
  
          
        manipulated_seive(N); 
       
        // pint all prime number less then N 
        for (int i=0; i<prime.size() && prime.get(i) <= N ; i++) 
            System.out.print(prime.get(i) + " "); 
    } 
} 

Output :

2 3 5 7 11

Illustration:

isPrime[0] = isPrime[1] = 0

After i = 2 iteration :
isPrime[]   [F, F, T, T, F, T, T, T]
SPF[]       [0, 0, 2, 0, 2, 0, 2, 0]
     index   0  1  2  3  4  5  6  7

After i = 3 iteration :
isPrime[]  [F, F, T, T, F, T, F, T, T, F ]
SPF[]      [0, 0, 2, 3, 2, 0, 2, 0, 0, 3 ]
  index     0  1  2  3  4  5  6  7  8  9

After i = 4 iteration :
isPrime[]  [F, F, T, T, F, T, F, T, F, F]
SPF[]      [0, 0, 2, 3, 2, 0, 2, 0, 2, 3]
  index     0  1  2  3  4  5  6  7  8  9

This article is contributed by Divyanshu Srivastava and Nishant Singh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

C++

Java

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