Not a kata suggestion.

This would be the best, yes, but there's one more, easier possibility. You can change the outer loop in a way that even if you keep both the loop and count, your solution would be reduced to O(n).

Getting back to my example: how many times do you need to count occurrences of 'a' in both strings to know the answer? How many times does your solution count occurrences of 'a'?

Let's say that you have following inputs:

s1 = "a" * 100_000
s2 = "a" * 100_000

What would be an expected answer? Would you have an idea why your solution is slow for this input, and how to make it faster?

i used js :(

hey man, can you help me out with my attempt
i have a question below you
thanks

Nice Kata

Fixed Java, should no longer accept O(n^2) solutions.

Hi;

In the Python version of this kata there are several random tests at the end of the test suite with over 100,000+ elements in each string (s1 and s2).

Your current approach is "correct" - the logic is correct - and will work OK for small input string lengths, but it will fail for such large tests.

The reason why is that you are iterating over every element of s1 for every element of s2 - in other words you are performing up to 100,000 * 100,000 operations which is a huge number.

Basically, you need to find a more efficient approach that does not require so many calculations.

std::string::npos is a size_t (kind of). It's just a number. How is a number too slow?

Here is a tip for C++:

This is slow:

std::string::npos

Why do you want to put a colon there? We need to separate two arguments (r and s) that are passed to the function.

Can somebody explain to me why the colon is not take into consideration when using "auto r" and "auto s"?