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# 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.

Input: n = 11
Output: true

Input: n = 15
Output: false

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

### 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```

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