Given a positive integer, check if the number is prime or not. A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. Examples of the first few prime numbers are {2, 3, 5, …}
Examples :
Input: n = 11
Output: true
Input: n = 15
Output: false
Input: n = 1
Output: false
School Method: A simple solution is to iterate through all numbers from 2 to n-1 and for every number check if it divides n. If we find any number that divides, we return false.
Below is the implementation of this method.
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 ;
}
int main()
{
isPrime(11) ? cout << " true\n" : cout << " false\n" ;
isPrime(15) ? cout << " true\n" : cout << " false\n" ;
return 0;
}
|
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 ;
}
public static void main(String args[])
{
if (isPrime( 11 ))
System.out.println( " true" );
else
System.out.println( " false" );
if (isPrime( 15 ))
System.out.println( " true" );
else
System.out.println( " false" );
}
}
|
Python3
def isPrime(n):
if n < = 1 :
return False
for i in range ( 2 , n):
if n % i = = 0 :
return False
return True
print ( "true" ) if isPrime( 11 ) else print ( "false" )
print ( "true" ) if isPrime( 14 ) else print ( "false" )
|
C#
using System;
namespace prime {
public class GFG {
public 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 ;
}
public static void Main()
{
if (isprime(11))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
if (isprime(15))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
}
}
}
|
PHP
<?php
function isPrime( $n )
{
if ( $n <= 1) return false;
for ( $i = 2; $i < $n ; $i ++)
if ( $n % $i == 0)
return false;
return true;
}
$tet = isPrime(11) ? " true\n" :
" false\n" ;
echo $tet ;
$tet = isPrime(15) ? " true\n" :
" false\n" ;
echo $tet ;
?>
|
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 ;
}
isPrime(11)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" );
isPrime(15)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" );
</script>
|
Time complexity: O(n)
Auxiliary Space: O(1)
Optimized School Method: We can do the following optimizations: Instead of checking till n, we can check till √n because a larger factor of n must be a multiple of a smaller factor that has been already checked. The implementation of this method is as follows:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1)
return false ;
for ( int i = 2; i <= sqrt (n); i++)
if (n % i == 0)
return false ;
return true ;
}
int main()
{
isPrime(11) ? cout << " true\n" : cout << " false\n" ;
isPrime(15) ? cout << " true\n" : cout << " false\n" ;
return 0;
}
|
Java
class GFG {
static boolean isPrime( int n)
{
if (n <= 1 )
return false ;
for ( int i = 2 ; i * i <= n; i++)
if (n % i == 0 )
return false ;
return true ;
}
public static void main(String args[])
{
if (isPrime( 11 ))
System.out.println( " true" );
else
System.out.println( " false" );
if (isPrime( 15 ))
System.out.println( " true" );
else
System.out.println( " false" );
}
}
|
Python3
import math
def isPrime(n):
if (n < = 1 ):
return False
for i in range ( 2 , int (math.sqrt(n)) + 1 ):
if (n % i = = 0 ):
return False
return True
print ( "true" ) if isPrime( 11 ) else print ( "false" )
print ( "true" ) if isPrime( 15 ) else print ( "false" )
|
C#
using System;
namespace prime {
public class GFG {
public static bool isprime( int n)
{
if (n <= 1)
return false ;
for ( int i = 2; i * i <= n; i++)
if (n % i == 0)
return false ;
return true ;
}
public static void Main()
{
if (isprime(11))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
if (isprime(15))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
}
}
}
|
Javascript
<script>
function isPrime(n)
{
if (n <= 1) return false ;
for (let i = 2; i*i <= n; i++)
if (n % i == 0)
return false ;
return true ;
}
if (isPrime(11))
document.write( " true" + "<br/>" );
else
document.write( " false" + "<br/>" );
if (isPrime(15))
document.write( " true" + "<br/>" );
else
document.write( " false" + "<br/>" );
</script>
|
Time Complexity: O(√n)
Auxiliary Space: O(1)
Another approach: It is based on the fact that all primes greater than 3 are of the form 6k ± 1, where k is any integer greater than 0. This is because all integers can be expressed as (6k + i), where i = −1, 0, 1, 2, 3, or 4. And note that 2 divides (6k + 0), (6k + 2), and (6k + 4) and 3 divides (6k + 3). So, a more efficient method is to test whether n is divisible by 2 or 3, then to check through all numbers of the form 6k ± 1 <= √n. This is 3 times faster than testing all numbers up to √n. (Source: wikipedia).
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n == 2 || n == 3)
return true ;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
}
int main()
{
isPrime(11) ? cout << " true\n" : cout << " false\n" ;
isPrime(15) ? cout << " true\n" : cout << " false\n" ;
return 0;
}
|
Java
class GFG {
static boolean isPrime( int n)
{
if (n == 2 || n == 3 )
return true ;
if (n <= 1 || n % 2 == 0 || n % 3 == 0 )
return false ;
for ( int i = 5 ; i * i <= n; i += 6 )
{
if (n % i == 0 || n % (i + 2 ) == 0 )
return false ;
}
return true ;
}
public static void main(String args[])
{
if (isPrime( 11 ))
System.out.println( " true" );
else
System.out.println( " false" );
if (isPrime( 15 ))
System.out.println( " true" );
else
System.out.println( " false" );
}
}
|
Python3
import math
def isPrime(n):
if n = = 2 or n = = 3 :
return True
elif n < = 1 or n % 2 = = 0 or n % 3 = = 0 :
return False
for i in range ( 5 , int (math.sqrt(n)) + 1 , 6 ):
if n % i = = 0 or n % (i + 2 ) = = 0 :
return False
return True
print (isPrime( 11 ))
print (isPrime( 20 ))
|
C#
using System;
class GFG {
static bool isPrime( int n)
{
if (n == 2 || n == 3)
return true ;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false ;
for ( int i = 5; i * i <= n; i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
}
public static void Main(String[] args)
{
if (isPrime(11))
Console.WriteLine( " true" );
else
Console.WriteLine( " false" );
if (isPrime(15))
Console.WriteLine( " true" );
else
Console.WriteLine( " false" );
}
}
|
Javascript
<script>
function isPrime(n)
{
if (n == 2 || n == 3)
return true ;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false ;
for (let i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
}
isPrime(11) ? document.write( " true" + "<br/>" ) : document.write( " false" + "<br/>" );
isPrime(15) ? document.write( " true" + "<br/>" ) : document.write( " false" + "<br/>" );
</script>
|
Time Complexity: O(√n)
Auxiliary Space: O(1)