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)
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