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: trueInput: n = 15
Output: falseInput: 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.
// A school method based C++ program to check if a // number is prime #include <bits/stdc++.h> using namespace std;
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 ;
} // Driver Program to test above function int main()
{ isPrime(11) ? cout << " true\n" : cout << " false\n" ;
isPrime(15) ? cout << " true\n" : cout << " false\n" ;
return 0;
} |
// A school method based JAVA program // to check if a number is prime class GFG {
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 ;
}
// Driver Program
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" );
}
} // This code is contributed // by Nikita Tiwari. |
# A school method based Python3 # program to check if a number # is prime 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
# Driver Program to test above function print ( "true" ) if isPrime( 11 ) else print ( "false" )
print ( "true" ) if isPrime( 14 ) else print ( "false" )
# This code is contributed by Smitha Dinesh Semwal |
// A optimized school method based C# // program to check if a number is prime using System;
namespace prime {
public class GFG {
public static bool isprime( int n)
{
// Corner cases
if (n <= 1)
return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
// Driver program
public static void Main()
{
if (isprime(11))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
if (isprime(15))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
}
} } // This code is contributed by Sam007 |
<?php // A school method based PHP // program to check if a number // is prime 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;
} // Driver Code $tet = isPrime(11) ? " true\n" :
" false\n" ;
echo $tet ;
$tet = isPrime(15) ? " true\n" :
" false\n" ;
echo $tet ;
// This code is contributed by m_kit ?> |
<script> // A school method based Javascript program to check if a // number is prime 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 ;
} // Driver Program to test above function isPrime(11)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" );
isPrime(15)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" );
// This code is contributed by Mayank Tyagi </script> |
true false
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:
// Optimised school method based C++ program to check if a // number is prime #include <bits/stdc++.h> using namespace std;
bool isPrime( int n)
{ // Corner case
if (n <= 1)
return false ;
// Check from 2 to square root of n
for ( int i = 2; i <= sqrt (n); i++)
if (n % i == 0)
return false ;
return true ;
} // Driver Program to test above function int main()
{ isPrime(11) ? cout << " true\n" : cout << " false\n" ;
isPrime(15) ? cout << " true\n" : cout << " false\n" ;
return 0;
} // This code is contributed by Vikash Sangai |
// Optimised school method based JAVA program // to check if a number is prime class GFG {
static boolean isPrime( int n)
{
// Corner case
if (n <= 1 )
return false ;
// Check from 2 to square root of n
for ( int i = 2 ; i * i <= n; i++)
if (n % i == 0 )
return false ;
return true ;
}
// Driver Program
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" );
}
} // This code is contributed by Vikash Sangai |
# Optimised school method based PYTHON program # to check if a number is prime # import the math module import math
# function to check whether the number is prime or not def isPrime(n):
# Corner case
if (n < = 1 ):
return False
# Check from 2 to square root of n
for i in range ( 2 , int (math.sqrt(n)) + 1 ):
if (n % i = = 0 ):
return False
return True
# Driver Program to test above function print ( "true" ) if isPrime( 11 ) else print ( "false" )
print ( "true" ) if isPrime( 15 ) else print ( "false" )
# This code is contributed by bhoomikavemula |
// Optimised school method based C# // program to check if a number is prime using System;
namespace prime {
public class GFG {
public static bool isprime( int n)
{
// Corner cases
if (n <= 1)
return false ;
for ( int i = 2; i * i <= n; i++)
if (n % i == 0)
return false ;
return true ;
}
// Driver program
public static void Main()
{
if (isprime(11))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
if (isprime(15))
Console.WriteLine( "true" );
else
Console.WriteLine( "false" );
}
} } // This code is contributed by Vikash Sangai |
<script> // JavaScript code for the above approach
function isPrime(n)
{
// Corner case
if (n <= 1) return false ;
// Check from 2 to square root of n
for (let i = 2; i*i <= n; i++)
if (n % i == 0)
return false ;
return true ;
}
// Driver Code
if (isPrime(11))
document.write( " true" + "<br/>" );
else
document.write( " false" + "<br/>" );
if (isPrime(15))
document.write( " true" + "<br/>" );
else
document.write( " false" + "<br/>" );
// This code is contributed by sanjoy_62. </script>
|
true false
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++ program to check the given number // is prime or not #include <bits/stdc++.h> using namespace std;
// Function to check if the given number // is prime or not. bool isPrime( int n)
{ if (n == 2 || n == 3)
return true ;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false ;
// To check through all numbers of the form 6k ± 1
for ( int i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
} // Driver Code int main()
{ isPrime(11) ? cout << " true\n" : cout << " false\n" ;
isPrime(15) ? cout << " true\n" : cout << " false\n" ;
return 0;
} |
// JAVA program to check the given number // is prime or not class GFG {
static boolean isPrime( int n)
{
// since 2 and 3 are prime
if (n == 2 || n == 3 )
return true ;
// if n<=1 or divided by 2 or 3 then it can not be prime
if (n <= 1 || n % 2 == 0 || n % 3 == 0 )
return false ;
// To check through all numbers of the form 6k ± 1
for ( int i = 5 ; i * i <= n; i += 6 )
{
if (n % i == 0 || n % (i + 2 ) == 0 )
return false ;
}
return true ;
}
// Driver Program
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" );
}
} // This code is contributed by Ujjwal Kumar Bhardwaj |
# Python program to check the given number # is prime or not # Function to check if the given number # is prime or not. 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
# To check through all numbers of the form 6k ± 1
# until i <= square root of n, with step value 6
for i in range ( 5 , int (math.sqrt(n)) + 1 , 6 ):
if n % i = = 0 or n % (i + 2 ) = = 0 :
return False
return True
# # Driver code print (isPrime( 11 ))
print (isPrime( 20 ))
# # This code is contributed by Harsh Master |
// C# program to check the given number // is prime or not using System;
class GFG {
static bool isPrime( int n)
{ // since 2 and 3 are prime
if (n == 2 || n == 3)
return true ;
// if n<=1 or divided by 2 or 3 then it can not be prime
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false ;
// To check through all numbers of the form 6k ± 1
for ( int i = 5; i * i <= n; i += 6)
{
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
} // Driver Program 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" );
} } // This code is contributed by Aman Kumar |
<script> // JavaScript program to check the given number
// is prime or not
// Function to check if the given number
// is prime or not.
function isPrime(n)
{
if (n == 2 || n == 3)
return true ;
if (n <= 1 || n % 2 == 0 || n % 3 == 0)
return false ;
// To check through all numbers of the form 6k ± 1
for (let i = 5; i * i <= n; i += 6) {
if (n % i == 0 || n % (i + 2) == 0)
return false ;
}
return true ;
}
// Driver Code
isPrime(11) ? document.write( " true" + "<br/>" ) : document.write( " false" + "<br/>" );
isPrime(15) ? document.write( " true" + "<br/>" ) : document.write( " false" + "<br/>" );
// This code is contributed by Pushpesh Raj.
</script> |
true false
Time Complexity: O(√n)
Auxiliary Space: O(1)