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Print all safe primes below N

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  • Difficulty Level : Medium
  • Last Updated : 23 Jun, 2022
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Given an integer N, the task is to print all safe primes below N safe primes. A safe prime is a prime number of the form (2 * p) + 1 where p is also a prime

The first few safe primes are 5, 7, 11, 23, 47, …

Examples:  

Input: N = 13 
Output: 5 7 11 
5 = 2 * 2 + 1 
7 = 2 * 3 + 1 
11 = 2 * 5 + 1

Input: N = 6 
Output: 5 7 
 

Approach: First pre-compute all the primes till N using Sieve of Eratosthenes and then starting from 2 check whether the current prime is also a safe prime. If yes then print it else skip to the next prime.

Below is the implementation of above approach:  

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to print first n safe primes
void printSafePrimes(int n)
{
    int prime[n + 1];
 
    // Initialize all entries of integer array
    // as 1. A value in prime[i] will finally
    // be 0 if i is Not a prime, else 1
    for (int i = 2; i <= n; i++)
        prime[i] = 1;
 
    // 0 and 1 are not primes
    prime[0] = prime[1] = 0;
 
    for (int p = 2; p * p <= n; p++) {
 
        // If prime[p] is not changed, then
        //  it is a prime
        if (prime[p] == 1) {
 
            // Update all multiples of p
            for (int i = p * 2; i <= n; i += p)
                prime[i] = 0;
        }
    }
 
    for (int i = 2; i <= n; i++) {
 
        // If i is prime
        if (prime[i] != 0) {
 
            // 2p + 1
            int temp = (2 * i) + 1;
 
            // If 2p + 1 is also a prime
            // then set prime[2p + 1] = 2
            if (temp <= n && prime[temp] != 0)
                prime[temp] = 2;
        }
    }
 
    for (int i = 5; i <= n; i++)
 
        // i is a safe prime
        if (prime[i] == 2)
            cout << i << " ";
}
 
// Driver code
int main()
{
    int n = 20;
    printSafePrimes(n);
    return 0;
}

Java




// Java implementation of the approach
class GFG{
     
    // Function to print first n safe primes
    static void printSafePrimes(int n)
    {
        int prime[] = new int [n + 1];
     
        // Initialize all entries of integer array
        // as 1. A value in prime[i] will finally
        // be 0 if i is Not a prime, else 1
        for (int i = 2; i <= n; i++)
            prime[i] = 1;
     
        // 0 and 1 are not primes
        prime[0] = prime[1] = 0;
     
        for (int p = 2; p * p <= n; p++)
        {
     
            // If prime[p] is not changed, then
            // it is a prime
            if (prime[p] == 1)
            {
     
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    prime[i] = 0;
            }
        }
     
        for (int i = 2; i <= n; i++)
        {
     
            // If i is prime
            if (prime[i] != 0)
            {
     
                // 2p + 1
                int temp = (2 * i) + 1;
     
                // If 2p + 1 is also a prime
                // then set prime[2p + 1] = 2
                if (temp <= n && prime[temp] != 0)
                    prime[temp] = 2;
            }
        }
     
        for (int i = 5; i <= n; i++)
     
            // i is a safe prime
            if (prime[i] == 2)
                System.out.print(i + " ");
    }
     
    // Driver code
    public static void main(String []args)
    {
        int n = 20;
        printSafePrimes(n);
    }
}
 
// This code is contributed by Ryuga

Python3




# Python 3 implementation of the approach
from math import sqrt
 
# Function to print first n safe primes
def printSafePrimes(n):
    prime = [0 for i in range(n + 1)]
 
    # Initialize all entries of integer
    # array as 1. A value in prime[i]
    # will finally be 0 if i is Not a
    # prime, else 1
    for i in range(2, n + 1):
        prime[i] = 1
 
    # 0 and 1 are not primes
    prime[0] = prime[1] = 0
 
    for p in range(2, int(sqrt(n)) + 1, 1):
         
        # If prime[p] is not changed,
        # then it is a prime
        if (prime[p] == 1):
             
            # Update all multiples of p
            for i in range(p * 2, n + 1, p):
                prime[i] = 0
     
    for i in range(2, n + 1, 1):
         
        # If i is prime
        if (prime[i] != 0):
             
            # 2p + 1
            temp = (2 * i) + 1
 
            # If 2p + 1 is also a prime
            # then set prime[2p + 1] = 2
            if (temp <= n and prime[temp] != 0):
                prime[temp] = 2
 
    for i in range(5, n + 1):
         
        # i is a safe prime
        if (prime[i] == 2):
            print(i, end = " ")
 
# Driver code
if __name__ == '__main__':
    n = 20
    printSafePrimes(n)
     
# This code is contributed by
# Sanjit_Prasad

C#




// C# implementation of the approach
using System;
 
class GFG{
     
    // Function to print first n safe primes
    static void printSafePrimes(int n)
    {
        int[] prime = new int [n + 1];
     
        // Initialize all entries of integer array
        // as 1. A value in prime[i] will finally
        // be 0 if i is Not a prime, else 1
        for (int i = 2; i <= n; i++)
            prime[i] = 1;
     
        // 0 and 1 are not primes
        prime[0] = prime[1] = 0;
     
        for (int p = 2; p * p <= n; p++)
        {
     
            // If prime[p] is not changed, then
            // it is a prime
            if (prime[p] == 1)
            {
     
                // Update all multiples of p
                for (int i = p * 2; i <= n; i += p)
                    prime[i] = 0;
            }
        }
     
        for (int i = 2; i <= n; i++)
        {
     
            // If i is prime
            if (prime[i] != 0)
            {
     
                // 2p + 1
                int temp = (2 * i) + 1;
     
                // If 2p + 1 is also a prime
                // then set prime[2p + 1] = 2
                if (temp <= n && prime[temp] != 0)
                    prime[temp] = 2;
            }
        }
     
        for (int i = 5; i <= n; i++)
     
            // i is a safe prime
            if (prime[i] == 2)
                Console.Write(i + " ");
    }
     
    // Driver code
    public static void Main()
    {
        int n = 20;
        printSafePrimes(n);
    }
}
 
// This code is contributed by Ita_c.

PHP




<?php
// PHP implementation of the approach
 
// Function to print first n safe primes
function printSafePrimes($n)
{
    $prime = array();
 
    // Initialize all entries of integer array
    // as 1. A value in prime[i] will finally
    // be 0 if i is Not a prime, else 1
    for ($i = 2; $i <= $n; $i++)
        $prime[$i] = 1;
 
    // 0 and 1 are not primes
    $prime[0] = $prime[1] = 0;
 
    for ($p = 2; $p * $p <= $n; $p++)
    {
 
        // If prime[p] is not changed,
        // then it is a prime
        if ($prime[$p] == 1)
        {
 
            // Update all multiples of p
            for ($i = $p * 2;
                 $i <= $n; $i += $p)
                $prime[$i] = 0;
        }
    }
 
    for ($i = 2; $i <= $n; $i++)
    {
 
        // If i is prime
        if ($prime[$i] != 0)
        {
 
            // 2p + 1
            $temp = (2 * $i) + 1;
 
            // If 2p + 1 is also a prime
            // then set prime[2p + 1] = 2
            if ($temp <= $n &&
                $prime[$temp] != 0)
                $prime[$temp] = 2;
        }
    }
 
    for ($i = 5; $i <= $n; $i++)
 
        // i is a safe prime
        if ($prime[$i] == 2)
            echo $i, " ";
}
 
// Driver code
$n = 20;
printSafePrimes($n);
 
// This code is contributed
// by aishwarya.27
?>

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to print first n safe primes
function printSafePrimes(n)
{
    let prime = new Array(n + 1);
   
    // Initialize all entries of integer array
    // as 1. A value in prime[i] will finally
    // be 0 if i is Not a prime, else 1
    for(let i = 2; i <= n; i++)
        prime[i] = 1;
   
    // 0 and 1 are not primes
    prime[0] = prime[1] = 0;
   
    for(let p = 2; p * p <= n; p++)
    {
         
        // If prime[p] is not changed, then
        // it is a prime
        if (prime[p] == 1)
        {
             
            // Update all multiples of p
            for(let i = p * 2; i <= n; i += p)
                prime[i] = 0;
        }
    }
   
    for(let i = 2; i <= n; i++)
    {
         
        // If i is prime
        if (prime[i] != 0)
        {
             
            // 2p + 1
            let temp = (2 * i) + 1;
   
            // If 2p + 1 is also a prime
            // then set prime[2p + 1] = 2
            if (temp <= n && prime[temp] != 0)
                prime[temp] = 2;
        }
    }
   
    for(let i = 5; i <= n; i++)
   
        // i is a safe prime
        if (prime[i] == 2)
            document.write(i + " ");
}
 
// Driver code
let n = 20;
printSafePrimes(n);
 
// This code is contributed by unknown2108
 
</script>

Output: 

5 7 11

 

Time complexity: O(nlog(logn)) same as Sieve of Eratosthenes

Auxiliary Space: O(n), since n extra space has been taken.


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