Given a number N, the task is to print all prime numbers less than or equal to N.
Examples:
Input: 7 Output: 2, 3, 5, 7 Input: 13 Output: 2, 3, 5, 7, 11, 13
Naive Approach: Iterate from 2 to N, and check for prime. If it is a prime number, print the number.
Below is the implementation of the above approach:
// C++ program to print all primes less than N #include <bits/stdc++.h> using namespace std;
// function check whether a number is prime or not bool isPrime( int n)
{ // Corner case
if (n <= 1)
return false ;
// Check from 2 to n-1
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
} // Function to print primes void printPrime( int n)
{ for ( int i = 2; i <= n; i++)
if (isPrime(i))
cout << i << " " ;
} // Driver Code int main()
{ int n = 7;
printPrime(n);
} |
// C program to print all primes less than N #include <stdbool.h> #include <stdio.h> // function check whether a number is prime or not bool isPrime( int n)
{ // Corner case
if (n <= 1)
return false ;
// Check from 2 to n-1
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
} // Function to print primes void printPrime( int n)
{ for ( int i = 2; i <= n; i++)
if (isPrime(i))
printf ( "%d " , i);
} // Driver Code int main()
{ int n = 7;
printPrime(n);
} // This code is contributed by Sania Kumari Gupta |
// Java program to print // all primes less than N class GFG {
// function check whether
// a number is prime or not
static boolean isPrime( int n)
{
// Corner case
if (n <= 1 )
return false ;
// Check from 2 to n-1
for ( int i = 2 ; i < n; i++)
if (n % i == 0 )
return false ;
return true ;
}
// Function to print primes
static void printPrime( int n)
{
for ( int i = 2 ; i <= n; i++) {
if (isPrime(i))
System.out.print(i + " " );
}
}
// Driver Code
public static void main(String[] args)
{
int n = 7 ;
printPrime(n);
}
} // This code is contributed // by ChitraNayal |
# Python3 program to print # all primes less than N # Function to check whether # a number is prime or not . def isPrime(n):
# Corner case
if n < = 1 :
return False
# check from 2 to n-1
for i in range ( 2 , n):
if n % i = = 0 :
return False
return True
# Function to print primes def printPrime(n):
for i in range ( 2 , n + 1 ):
if isPrime(i):
print (i, end = " " )
# Driver code if __name__ = = "__main__" :
n = 7
# function calling
printPrime(n)
# This code is contributed # by Ankit Rai |
// C# program to print // all primes less than N using System;
class GFG
{ // function check whether // a number is prime or not static bool isPrime( int n)
{ // Corner case
if (n <= 1)
return false ;
// Check from 2 to n-1
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
} // Function to print primes static void printPrime( int n)
{ for ( int i = 2; i <= n; i++)
{ if (isPrime(i))
Console.Write(i + " " );
} } // Driver Code public static void Main()
{ int n = 7;
printPrime(n);
} } // This code is contributed // by ChitraNayal |
<?php // PHP program to print // all primes less than N // function check whether // a number is prime or not function isPrime( $n )
{ // Corner case
if ( $n <= 1)
return false;
// Check from 2 to n-1
for ( $i = 2; $i < $n ; $i ++)
if ( $n % $i == 0)
return false;
return true;
} // Function to print primes function printPrime( $n )
{ for ( $i = 2; $i <= $n ; $i ++)
{
if (isPrime( $i ))
echo $i . " " ;
}
} // Driver Code $n = 7;
printPrime( $n );
// This code is contributed // by ChitraNayal ?> |
<script> // Javascript program to print all primes // less than N // function check whether a number // is prime or not function isPrime(n)
{ // Corner case
if (n <= 1)
return false ;
// Check from 2 to n-1
for (let i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
} // Function to print primes function printPrime(n)
{ for (let i = 2; i <= n; i++) {
if (isPrime(i))
document.write(i + " " );
}
} // Driver Code let n = 7;
printPrime(n);
// This code is contributed by Mayank Tyagi </script> |
Output:
2 3 5 7
Time Complexity: O(N * N)
Auxiliary Space: O(1)
A better approach is based on the fact that one of the divisors must be smaller than or equal to ?n. So we check for divisibility only till ?n.
// C++ program to print all primes // less than N #include <bits/stdc++.h> using namespace std;
// function check whether a number // is prime or not bool isPrime( int n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to print primes void printPrime( int n)
{ for ( int i = 2; i <= n; i++) {
if (isPrime(i))
cout << i << " " ;
}
} // Driver Code int main()
{ int n = 7;
printPrime(n);
} |
// Java program to print // all primes less than N import java.io.*;
class GFG
{ // function check // whether a number // is prime or not static boolean isPrime( int n)
{ // Corner cases
if (n <= 1 )
return false ;
if (n <= 3 )
return true ;
// This is checked so
// that we can skip
// middle five numbers
// in below loop
if (n % 2 == 0 ||
n % 3 == 0 )
return false ;
for ( int i = 5 ;
i * i <= n; i = i + 6 )
if (n % i == 0 ||
n % (i + 2 ) == 0 )
return false ;
return true ;
} // Function to print primes static void printPrime( int n)
{ for ( int i = 2 ; i <= n; i++)
{
if (isPrime(i))
System.out.print(i + " " );
}
} // Driver Code public static void main (String[] args)
{ int n = 7 ;
printPrime(n);
} } // This code is contributed // by anuj_67. |
// C# program to print // all primes less than N using System;
class GFG
{ // function check // whether a number // is prime or not static bool isPrime( int n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// This is checked so
// that we can skip
// middle five numbers
// in below loop
if (n % 2 == 0 ||
n % 3 == 0)
return false ;
for ( int i = 5;
i * i <= n; i = i + 6)
if (n % i == 0 ||
n % (i + 2) == 0)
return false ;
return true ;
} // Function to print primes static void printPrime( int n)
{ for ( int i = 2; i <= n; i++)
{
if (isPrime(i))
Console.Write(i + " " );
}
} // Driver Code public static void Main ()
{ int n = 7;
printPrime(n);
} } // This code is contributed // by ChitraNayal |
# function to check if the number is # prime or not def isPrime(n) :
# Corner cases
if (n < = 1 ) :
return False
if (n < = 3 ) :
return True
# This is checked so that we can skip
# middle five numbers in below loop
if (n % 2 = = 0 or n % 3 = = 0 ) :
return False
i = 5
while (i * i < = n) :
if (n % i = = 0 or n % (i + 2 ) = = 0 ) :
return False
i = i + 6
return True # print all prime numbers # less than equal to N def printPrime(n):
for i in range ( 2 , n + 1 ):
if isPrime(i):
print (i, end = " " )
n = 7 printPrime(n) |
<script> // Javascript program to print all primes // less than N // function check whether a number // is prime or not function isPrime(n)
{ // Corner cases
if (n <= 1)
return false ;
if (n <= 3)
return true ;
// This is checked so that we can skip
// middle five numbers in below loop
if (n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5; i * i <= n; i = i + 6)
if (n % i == 0 || n % (i + 2) == 0)
return false ;
return true ;
} // Function to print primes function printPrime(n)
{ for (let i = 2; i <= n; i++) {
if (isPrime(i))
document.write(i + " " );
}
} // Driver Code let n = 7; printPrime(n); // This code is contributed by subhammahato348. </script> |
<?php // PHP program to print // all primes less than N // function check whether // a number is prime or not function isPrime( $n )
{ // Corner cases
if ( $n <= 1)
return false;
if ( $n <= 3)
return true;
// This is checked so that
// we can skip middle five
// numbers in below loop
if ( $n % 2 == 0 || $n % 3 == 0)
return false;
for ( $i = 5;
$i * $i <= $n ; $i = $i + 6)
if ( $n % $i == 0 ||
$n % ( $i + 2) == 0)
return false;
return true;
} // Function to print primes function printPrime( $n )
{ for ( $i = 2; $i <= $n ; $i ++)
{
if (isPrime( $i ))
echo $i . " " ;
}
} // Driver Code $n = 7;
printPrime( $n );
// This code is contributed // by ChitraNayal ?> |
2 3 5 7
Time Complexity: O(N3/2)
Auxiliary Space: O(1)
The best solution is to use Sieve of Eratosthenes. The time complexity is O(N * loglog(N))