Python Program for Product of unique prime factors of a number

Given a number n, we need to find the product of all of its unique prime factors. Prime factors: It is basically a factor of the number that is a prime number itself. Examples:

Input: num = 10
Output: Product is 10
Explanation:
Here, the input number is 10 having only 2 prime factors and they are 5 and 2.
And hence their product is 10.

Input : num = 25
Output: Product is 5
Explanation:
Here, for the input to be 25  we have only one unique prime factor i.e 5.
And hence the required product is 5.

Method 1 (Simple) Using a loop from i = 2 to n and check if i is a factor of n then check if i is prime number itself if yes then store product in product variable and continue this process till i = n. 

Python3

# Python program to find sum of given
# series.


def productPrimeFactors(n):
    product = 1

    for i in range(2, n+1):
        if (n % i == 0):
            isPrime = 1

            for j in range(2, int(i/2 + 1)):
                if (i % j == 0):
                    isPrime = 0
                    break

            # condition if \'i\' is Prime number
            # as well as factor of num
            if (isPrime):
                product = product * i

    return product


# main()
n = 44
print(productPrimeFactors(n))

# Contributed by _omg

Output:

22

Time complexity: O(n^2/2)
Auxiliary space: O(1)

Method 2 (Efficient) : The idea is based on Efficient program to print all prime factors of a given number

Python3

# Python program to find product of
# unique prime factors of a number

import math

def productPrimeFactors(n):
    product = 1

    # Handle prime factor 2 explicitly so that
    # can optimally handle other prime factors.
    if (n % 2 == 0):
        product *= 2
        while (n%2 == 0):
            n = n/2

    # n must be odd at this point. So we can
    # skip one element (Note i = i +2)
    for i in range (3, int(math.sqrt(n)), 2):
        # While i divides n, print i and
        # divide n
        if (n % i == 0):
            product = product * i
            while (n%i == 0):
                n = n/i

    # This condition is to handle the case when n
    # is a prime number greater than 2
    if (n > 2):
        product = product * n

    return product

# main()
n = 44
print (int(productPrimeFactors(n)))

# Contributed by _omg

Output:

22

Time complexity: O(sqrt(n)), where n is the input number.
Auxiliary space: O(1), as the program only uses a constant amount of memory to store the product and the loop variables.

Please refer complete article on Product of unique prime factors of a number for more details!