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# Python Program to Check Prime Number

• Difficulty Level : Easy
• Last Updated : 23 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))

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`

Output

```True
False```

Time Complexity: O(sqrt(n))
Auxiliary Space: O(1)

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