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
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.
C++
#include <bits/stdc++.h>
using namespace std;
bool isPrime( int n)
{
if (n <= 1) return false ;
for ( int i=2; i<n; i++)
if (n%i == 0)
return false ;
return true ;
}
int main()
{
isPrime(11)? cout << " true\n" : cout << " false\n" ;
isPrime(15)? cout << " true\n" : cout << " false\n" ;
return 0;
}
|
Java
class GFG {
static boolean isPrime( int n)
{
if (n <= 1 ) return false ;
for ( int i = 2 ; i < n; i++)
if (n % i == 0 )
return false ;
return true ;
}
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" );
}
}
|
Python3
def isPrime(n):
if n < = 1 :
return False
for i in range ( 2 , n):
if n % i = = 0 :
return False ;
return True
print ( "true" ) if isPrime( 11 ) else print ( "false" )
print ( "true" ) if isPrime( 14 ) else print ( "false" )
|
C#
using System;
namespace prime
{
public class GFG
{
public static bool isprime( int n)
{
if (n <= 1) return false ;
for ( int i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
public static void Main()
{
if (isprime(11)) Console.WriteLine( "true" );
else Console.WriteLine( "false" );
if (isprime(15)) Console.WriteLine( "true" );
else Console.WriteLine( "false" );
}
}
}
|
PHP
<?php
function isPrime( $n )
{
if ( $n <= 1) return false;
for ( $i = 2; $i < $n ; $i ++)
if ( $n % $i == 0)
return false;
return true;
}
$tet = isPrime(11) ? " true\n" :
" false\n" ;
echo $tet ;
$tet = isPrime(15) ? " true\n" :
" false\n" ;
echo $tet ;
?>
|
Javascript
<script>
function isPrime(n)
{
if (n <= 1) return false ;
for (let i = 2; i < n; i++)
if (n % i == 0)
return false ;
return true ;
}
isPrime(11)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" );
isPrime(15)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" );
</script>
|
Output :
true
false
Time complexity of this solution is O(n)
Optimized School Method
We can do following optimizations:
- 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.
- 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)