# Python Program to Check Prime Number

• Difficulty Level : Easy
• Last Updated : 14 Mar, 2023

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, â€¦.}. ```

## 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"``)`

Output

`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 âˆšn 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 âˆšn )

## 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"``)`

Output

`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))`

Output

`True`

Time complexity: O(sqrt(n))
Auxiliary space: O(sqrt(n))

## 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``))`

Output

`True`

Time complexity: O(sqrt(n))
Auxiliary space: O(sqrt(n))

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