Thanks for saying why it works.

the fun fact, is you used the same exact code as mine... even the name of vars xD or the structer, I really find it strage we both used it this way

exactly, if not zero it is false? while if S1 is larger it should be true
because it means it contianed all the letters required and more

so I am a little bit confused, what if s1 has more length, more letter for each alphabet, it should not be 0?

I forgot to say why I assumed the other solutions as 2n,
well because they compare on every iteration both elements
(to make you realize how bad is it, just assume the array is sorted try 10 numbers array)
see how many swaps and how many comparisons for such algorithm
to build an intuition on how bad it is.

if list are worse then I may suggest sets?
and btw, the idea I meant wasn't about dict or list more than about why counting frequencies of words when all we need to do is terminating when the first repeatition occurs.

I know there are many solutions however why do you want to use dictionaries? you are counting the frequencies of the letters which is not needed
you can terminate right away when a repeat is occur, I did that in my solution you can check it if you like however, you can solve this with list, tuples dict and sets
I would say list would be the best since there is no need to use a dictionary.

Can you give me a solution that is than mine, thanks in advance
And since 2 people already said it.
I will say it in advance if you read my code closely my solution takes O(n+sqrt(n)) which is faster
than O(2n)
so If you know any solution that is faster than mine I am very happy to hear from you.

My solution is faster, I can't understand how the best solution is being slow....

even thought it's reading complexity is almost zero while mine... I had to put documentation to make it readable haha