A prime is a natural number greater than 1 that has no positive divisors other than 1 and itself. For example 2, 3, 5, 7, 11,….. are prime numbers.
In this article, we will learn how to write a prime number program in Java, when the input given is a Positive number.
Methods to Write Prime Number Program in Java
For checking a prime number in Java there is no formulae available but there are a few methods available to check if a number is Prime or not. There are a few methods to check if a number is prime or not as mentioned below:
1. Simple Program to Check Prime in Java
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 Java program to implement the above approach:
// Java Program to demonstrate // Brute Force Method // 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" );
}
} |
true false
The complexity of the above method
Time complexity: O(n)
Space complexity: O(1)
2. Improved Method in Java to Check Prime
In this method, the check is done from 2 to n/2 as a number is not divisible by more than half its value.
Below is a Java program to implement the approach:
// JAVA program to demonstrate // Improved method // to check if a number is prime import java.util.Scanner;
// Driver Class class GFG {
static boolean isPrime( int n)
{
// Corner case
if (n <= 1 )
return false ;
// Check from 2 to n/2
for ( int i = 2 ; i <= n / 2 ; 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" );
}
} |
true false
Complexity of the above method
Time complexity: O(N)
Space complexity: O(1)
3. Optimized Java Code for Prime Number
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.
Below is the Java program to implement the above approach:
// Java Program to demonstrate // Optimized method // to check if a number is prime import java.util.Scanner;
class GFG {
static boolean isPrime( int n)
{
// Corner case
if (n <= 1 )
return false ;
// Check from 2 to sqrt(n)
for ( int i = 2 ; i <= Math.sqrt(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" );
}
} |
true false
Complexity of the above method
Time complexity: O(√n)
Space complexity: O(1)
4. Most Optimized Method
The algorithm can be improved further by observing that all primes are of the form 6k ± 1, with the exception of 2 and 3. This is because all integers can be expressed as (6k + i) for some integer k and for i = -1, 0, 1, 2, 3, or 4.
Note: 2 divides (6k + 0), (6k + 2), (6k + 4)
3 divides (6k + 3)
So, a more efficient method is to test if n is divisible by 2 or 3, then to check through all the numbers of form 6k ± 1 ≤ √n. This approach is 3 times faster than testing all numbers up to √n.
Below is the Java program to implement the above approach:
// JAVA program to demonstrate // Optimized method based // to check if a number is prime import java.util.Scanner;
class GFG {
static boolean isPrime( int n)
{
// Corner case
if (n <= 1 )
return false ;
if (n == 2 || n == 3 )
return true ;
if (n % 2 == 0 || n % 3 == 0 )
return false ;
for ( int i = 5 ; i <= Math.sqrt(n); i = 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" );
}
} |
true false
Complexity of the above method
Time complexity: O(√n)
Space complexity: O(1)
To know more, please refer to the complete article – Prime Numbers