I like the trick of usingnot to reverse the value at the index, but I'm not convinced that the way you calculate indices here is easier to read than using the% operator.

Personally, I think the following is easier to understand:

foriinrange(1,101):# If the index is a multiple of the step, reverse the# hat status of the cat at that indexifi%step==0:cats[i-1]=notcats[i-1]

Granted, that solution requires an extra line of code, but IMO it's a lot easier to digest than first computing the number of multiples of step there are between 1 and 100, then computing those multiples in the indices.

I agree with you, the way I wrote this part is clearly not easy to read.

I have a new suggestion as I don't like the fact that we loop over each cat and test them when we can know from the start exactly which ones we need to visit.

# For each step size we calculate how many cats we are going to visitcats_to_visit=100//step_sizeforiinrange (1,cats_to_visit+1):# We reverse the hat status for each cat we visitcats[i*step]=notcats[i*step]

What do you think?

If you agree, I am a beginner with Github, should I directly edit my file with this?

+1 on this part of the solution. The statement of the problem says:

Write a program that simply outputs which cats have hats at the end.

I think the other solutions actually don't do a good job of this. They print a list of indices of cats that have hats.

Printing something like what you have here seems more in line with what the challenge asks for. 👍