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# Java program to check if a number is prime or not

• Difficulty Level : Medium
• Last Updated : 18 Oct, 2018

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 first few prime numbers are {2, 3, 5,

Examples :

```Input:  n = 11
Output: true

Input:  n = 15
Output: false

Input:  n = 1
Output: false
```

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

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.

 `// 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"``);``    ``}``}`
Output:
```true
false
```

Time complexity of this solution is O(n)

Optimized School Method
We can do following optimizations:

1. Instead of checking till n, we can check till √n because a larger factor of n must be a multiple of smaller factor that has been already checked.
2. 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; 2 divides (6k + 0), (6k + 2), (6k + 4); and 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. (Source: wikipedia)
 `// A optimized school method based Java``// program to check if a number is prime``import` `java.io.*;`` ` `class` `GFG {`` ` `    ``static` `boolean` `isPrime(``int` `n)``    ``{``        ``// Corner cases``        ``if` `(n <= ``1``)``            ``return` `false``;``        ``if` `(n <= ``3``)``            ``return` `true``;`` ` `        ``// This is checked so that we can skip``        ``// middle five numbers in below loop``        ``if` `(n % ``2` `== ``0` `|| n % ``3` `== ``0``)``            ``return` `false``;`` ` `        ``for` `(``int` `i = ``5``; i * i <= 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
```

Time complexity of this solution is O(√n)

Main Article : Primality Test | Set 1 (Introduction and School Method)