Print all prime numbers less than or equal to N
Last Updated :
13 Mar, 2023
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++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1)
return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
void printPrime( int n)
{
for ( int i = 2; i <= n; i++)
if (isPrime(i))
cout << i << " " ;
}
int main()
{
int n = 7;
printPrime(n);
}
|
C
#include <stdbool.h>
#include <stdio.h>
bool isPrime( int n)
{
if (n <= 1)
return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
void printPrime( int n)
{
for ( int i = 2; i <= n; i++)
if (isPrime(i))
printf ( "%d " , i);
}
int main()
{
int n = 7;
printPrime(n);
}
|
Java
class GFG {
static boolean isPrime( int n)
{
if (n <= 1 )
return false ;
for ( int i = 2 ; i < n; i++)
if (n % i == 0 )
return false ;
return true ;
}
static void printPrime( int n)
{
for ( int i = 2 ; i <= n; i++) {
if (isPrime(i))
System.out.print(i + " " );
}
}
public static void main(String[] args)
{
int n = 7 ;
printPrime(n);
}
}
|
Python3
def isPrime(n):
if n < = 1 :
return False
for i in range ( 2 , n):
if n % i = = 0 :
return False
return True
def printPrime(n):
for i in range ( 2 , n + 1 ):
if isPrime(i):
print (i, end = " " )
if __name__ = = "__main__" :
n = 7
printPrime(n)
|
C#
using System;
class GFG
{
static bool isPrime( int n)
{
if (n <= 1)
return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
static void printPrime( int n)
{
for ( int i = 2; i <= n; i++)
{
if (isPrime(i))
Console.Write(i + " " );
}
}
public static void Main()
{
int n = 7;
printPrime(n);
}
}
|
PHP
<?php
function isPrime( $n )
{
if ( $n <= 1)
return false;
for ( $i = 2; $i < $n ; $i ++)
if ( $n % $i == 0)
return false;
return true;
}
function printPrime( $n )
{
for ( $i = 2; $i <= $n ; $i ++)
{
if (isPrime( $i ))
echo $i . " " ;
}
}
$n = 7;
printPrime( $n );
?>
|
Javascript
<script>
function isPrime(n)
{
if (n <= 1)
return false ;
for (let i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
function printPrime(n)
{
for (let i = 2; i <= n; i++) {
if (isPrime(i))
document.write(i + " " );
}
}
let n = 7;
printPrime(n);
</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++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
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 ;
}
void printPrime( int n)
{
for ( int i = 2; i <= n; i++) {
if (isPrime(i))
cout << i << " " ;
}
}
int main()
{
int n = 7;
printPrime(n);
}
|
Java
import java.io.*;
class GFG
{
static boolean isPrime( int n)
{
if (n <= 1 )
return false ;
if (n <= 3 )
return true ;
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 ;
}
static void printPrime( int n)
{
for ( int i = 2 ; i <= n; i++)
{
if (isPrime(i))
System.out.print(i + " " );
}
}
public static void main (String[] args)
{
int n = 7 ;
printPrime(n);
}
}
|
C#
using System;
class GFG
{
static bool isPrime( int n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
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 ;
}
static void printPrime( int n)
{
for ( int i = 2; i <= n; i++)
{
if (isPrime(i))
Console.Write(i + " " );
}
}
public static void Main ()
{
int n = 7;
printPrime(n);
}
}
|
Python3
def isPrime(n) :
if (n < = 1 ) :
return False
if (n < = 3 ) :
return True
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
def printPrime(n):
for i in range ( 2 , n + 1 ):
if isPrime(i):
print (i, end = " " )
n = 7
printPrime(n)
|
Javascript
<script>
function isPrime(n)
{
if (n <= 1)
return false ;
if (n <= 3)
return true ;
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 printPrime(n)
{
for (let i = 2; i <= n; i++) {
if (isPrime(i))
document.write(i + " " );
}
}
let n = 7;
printPrime(n);
</script>
|
PHP
<?php
function isPrime( $n )
{
if ( $n <= 1)
return false;
if ( $n <= 3)
return true;
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 printPrime( $n )
{
for ( $i = 2; $i <= $n ; $i ++)
{
if (isPrime( $i ))
echo $i . " " ;
}
}
$n = 7;
printPrime( $n );
?>
|
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))
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