Java program to check if a number is prime or not

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. For example 2, 3, 5, 7, 11,….. are prime numbers.

Examples

Input: n = 11
Output: true

Input: n = 15
Output: false

Simple 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 Java program to implement the above approach:

Java

// 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");
    }
}

Output

 true
 false

Time complexity:  O(n) 

Improved Method

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

// 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");
    }
}

Output

 true
 false

Optimized Method

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

// 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");
    }
}

Output

 true
 false

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

// 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");
    }
}

Output

 true
 false