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I am glad I could help!

Just working on VSCode :)

The divisor is not incremented to the next prime factor "technically" but the divisor is only added to the list of prime factors only if it is a prime factor.

@@kaifk yes you are right, i know this while typing the comment, but i just need a person to reply me... thanks

Can you post your code and logs so we can have a look?

If you're looking to remove duplicate prime factors you have a few approaches. You can conditionally append factors to the list of prime factors i.e. check if the number is already in the list and only if it's not, add it to the list. You could also return the list as a set by calling set(name_of_list) and that will remove all duplicates. There are more ways to do it but these are 2 simple methods.

I believe the time complexity in Big O notation is O(n).

Check this out: www.romannumerals.org/blog/which-is-the-biggest-number-in-roman-numerals-6

@@kaifk thanks, managed to get it to work now 😁

Thanks for the advice I'll try that out for my next video :)

def prime_factor(n): """ this function returns the prime factors of a number """ prime_factors = [] divisor = 2 while divisor <= n: if n%divisor == 0: prime_factors.append(divisor) n = n/divisor else: divisor += 1 return prime_factor(n) n = int(input("Enter a number:")) print("Prime factors of {0} are:".format(n, prime_factor(n)))

The point of that is once you've found a factor you need to divide the original number by the factor to get the next otherwise your loop would never end and you'll keep getting the same prime factor.

Your return statement is wrong, you want to return `prime_factors` not `prime_factor(n)`

@@kaifk yes i realized it right after i read ur comment and went through the code. Thanks.

@@kaifk Could u help me with following:- def multiplication_table(num, i): """ this function returns the multiplication table of a number """ for i in range (1, 11): if num >= 0: product = num*i return product num = int(input("Enter positive no.")) print("{0} * {1} = {2}". format(num, i, multiplication_table(num, i))) error: name 'i' is not defined

Thank you so much

Hi Hamid, these are great suggestions, I will keep the shorthand syntax in mind for my next video and the suggestion of adding 2 after the divisor has reached 3 is a great little optimization. Thanks so much :)

@@kaifk Your content is great, please keep them coming!

Wouldn’t incrementing the divisor by 2 each time skip some factors?? Since on the first iteration, divisor is 2, incrementing by 2 would skip 3, no?? I need clarification..

Thank you for the feedback, I'll take it on board for the next video :)

I'm not sure what you mean but thank you 😅

Thank you so much!

@@kaifk You haven't posted any videos lately, I hope things are going well. 🙌🏻

Thanks for checking in, things have been busy lately I'll be back with some content very soon :) (I really appreciate it, genuinely my favorite comment)

Thank you so much!

Have you tried it? It would be fun to see what happens (how long does the number have to be till your CPU can't handle it) I'd imagine higher than 11 since you're dividing by at least 2 each iteration 🤷

The code in the video is linked in the description.