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++
// A school method based C++ program to check if a // number is prime #include <bits/stdc++.h> using namespace std; bool 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 to test above function int main() { isPrime(11)? cout << " true\n" : cout << " false\n" ; isPrime(15)? cout << " true\n" : cout << " false\n" ; return 0; } |
Java
// 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" ); } } // This code is contributed // by Nikita Tiwari. |
Python3
# A school method based Python3 # program to check if a number # is prime def isPrime(n): # Corner case if n < = 1 : return False # Check from 2 to n-1 for i in range ( 2 , n): if n % i = = 0 : return False ; return True # Driver Program to test above function print ( "true" ) if isPrime( 11 ) else print ( "false" ) print ( "true" ) if isPrime( 14 ) else print ( "false" ) # This code is contributed by Smitha Dinesh Semwal |
C#
// A optimized school method based C# // program to check if a number is prime using System; namespace prime { public class GFG { public static bool isprime( int n) { // Corner cases if (n <= 1) return false ; for ( int i = 2; i < n; i++) if (n % i == 0) return false ; return true ; } // Driver program public static void Main() { if (isprime(11)) Console.WriteLine( "true" ); else Console.WriteLine( "false" ); if (isprime(15)) Console.WriteLine( "true" ); else Console.WriteLine( "false" ); } } } // This code is contributed by Sam007 |
PHP
<?php // A school method based PHP // program to check if a number // is prime function isPrime( $n ) { // Corner case if ( $n <= 1) return false; // Check from 2 to n-1 for ( $i = 2; $i < $n ; $i ++) if ( $n % $i == 0) return false; return true; } // Driver Code $tet = isPrime(11) ? " true\n" : " false\n" ; echo $tet ; $tet = isPrime(15) ? " true\n" : " false\n" ; echo $tet ; // This code is contributed by m_kit ?> |
Javascript
<script> // A school method based Javascript program to check if a // number is prime function isPrime(n) { // Corner case if (n <= 1) return false ; // Check from 2 to n-1 for (let i = 2; i < n; i++) if (n % i == 0) return false ; return true ; } // Driver Program to test above function isPrime(11)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" ); isPrime(15)? document.write( " true" + "<br>" ): document.write( " false" + "<br>" ); // This code is contributed by Mayank Tyagi </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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