Given a positive integer N, The task is to write a Python program to check if the number is prime or not.
Definition: A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. The first few prime numbers are {2, 3, 5, 7, 11, ….}.
Examples :
Input: n = 11
Output: trueInput: n = 15
Output: falseInput: n = 1
Output: false
The idea to solve this problem is to iterate through all the numbers starting from 2 to (N/2) using a for loop and for every number check if it divides N. If we find any number that divides, we return false. If we did not find any number between 2 and N/2 which divides N then it means that N is prime and we will return True.
Below is the Python program to check if a number is prime:
C
// C program for // the above approach #include <stdio.h> int main() { // Given number int n = 11; // checking the given number // whether it is 1 or not if (n == 1) { printf ( "%d is not a prime number" , n); } else { int f = 0; // iterate from 2 to n/2 for ( int i = 2; i <= (n / 2); i++) { // If n is divisible by any number between // 2 and n/2, it is not prime if (n % 2 == 0) { f = 1; // break out of for loop as // it is not prime break ; } } if (f == 1) { printf ( "%d is not a prime number" , n); } else { printf ( "%d is a prime number" , n); } } return 0; } // This Code is Contributed by // Murarishetty Santhosh Charan |
Python3
# Python program to check if # given number is prime or not num = 11 # If given number is greater than 1 if num > 1 : # Iterate from 2 to n / 2 for i in range ( 2 , int (num / 2 ) + 1 ): # If num is divisible by any number between # 2 and n / 2, it is not prime if (num % i) = = 0 : print (num, "is not a prime number" ) break else : print (num, "is a prime number" ) else : print (num, "is not a prime number" ) |
C++
// C++ program for // the above approach #include <bits/stdc++.h> using namespace std; int main() { // Given number int n=11; // checking the given number // whether it is 1 or not if (n==1) { cout<<n<< " is not a prime number" ; } else { int f=0; // iterate from 2 to n/2 for ( int i=2;i<=(n/2);i++) { // If n is divisible by any number between // 2 and n/2, it is not prime if (n%2==0) { f=1; // break out of for loop as // it is not prime break ; } } if (f==1) { cout<<n<< " is not a prime number" ; } else { cout<<n<< " is a prime number" ; } } return 0; } // This code is contributed by // Murarishetty Santhosh Charan |
11 is a prime number
Optimized Method
We can do the following optimizations:
- 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.
- 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)
Python3
# A optimized school method based # Python3 program to check # if a number is prime def isPrime(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 or n % 3 = = 0 ) : return False i = 5 while (i * i < = n) : if (n % i = = 0 or n % (i + 2 ) = = 0 ) : return False i = i + 6 return True # Driver Program if (isPrime( 11 )) : print ( " true" ) else : print ( " false" ) if (isPrime( 15 )) : print ( " true" ) else : print ( " false" ) # This code is contributed # by Nikita Tiwari. |
Output:
true false
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