Python Program to Check Prime Number
Given a positive integer N, The task is to write a Python program to check if the number is Prime or not in Python.
Examples:
Input: n = 11 Output: True Input: n = 1 Output: False Explanation: 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, ….}.
Prime Number Program in Python
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.
Python3
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" ) |
11 is a prime number
Time complexity: O(n)
Auxiliary space: O(1)
Fastest Algorithm to Find Prime Numbers
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. Now let’s see the code for the first optimization method ( i.e. checking till √n )
Python3
from math import sqrt # n is the number to be check whether it is prime or not n = 1 # this flag maintains status whether the n is prime or not prime_flag = 0 if (n > 1 ): for i in range ( 2 , int (sqrt(n)) + 1 ): if (n % i = = 0 ): prime_flag = 1 break if (prime_flag = = 0 ): print ( "True" ) else : print ( "False" ) else : print ( "False" ) |
False
Time complexity: O(sqrt(n))
Auxiliary space: O(1)
Check Prime Numbers Using recursion
We can also find the number prime or not using recursion. We can use the exact logic shown in method 2 but in a recursive way.
Python3
from math import sqrt def Prime(number,itr): #prime function to check given number prime or not if itr = = 1 : #base condition return True if number % itr = = 0 : #if given number divided by itr or not return False if Prime(number,itr - 1 ) = = False : #Recursive function Call return False return True num = 13 itr = int (sqrt(num) + 1 ) print (Prime(num,itr)) |
True
Time complexity: O(sqrt(n))
Auxiliary space: O(sqrt(n))
Check the Prime Trial Division Method
Python3
def is_prime_trial_division(n): # Check if the number is less than or equal to 1, return False if it is if n < = 1 : return False # Loop through all numbers from 2 to # the square root of n (rounded down to the nearest integer) for i in range ( 2 , int (n * * 0.5 ) + 1 ): # If n is divisible by any of these numbers, return False if n % i = = 0 : return False # If n is not divisible by any of these numbers, return True return True # Test the function with n = 11 print (is_prime_trial_division( 11 )) |
True
Time complexity: O(sqrt(n))
Auxiliary space: O(sqrt(n))
RECOMMENDED ARTICLE – Analysis of Different Methods to find Prime Number in Python
Using a while loop to check divisibility:
Approach:
Initialize a variable i to 2.
While i squared is less than or equal to n, check if n is divisible by i.
If n is divisible by i, return False.
Otherwise, increment i by 1.
If the loop finishes without finding a divisor, return True.
Python3
import math def is_prime(n): if n < 2 : return False i = 2 while i * i < = n: if n % i = = 0 : return False i + = 1 return True print (is_prime( 11 )) # True print (is_prime( 1 )) # False |
True False
Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)
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