Woops the formatting is botched, my bad

I only saw your reply when I posted mine and the page refreshed. At least you got a second confirmation.

Is 44 your answer or the expected answer ? Because 39 seems correct to me.

• First do step 1, 2 and 4 (takes 13 minutes)
• Then do step 6 in the background (will last for 21 more minutes)
• Do steps 3 and 5 while the carrots boil (doesn't take time as step 6 is longer)
• Finish with steps 7 and 8 (takes 5 minutes)

13 + 21 + 5 = 39

In assert_eq!(...), when the two values compare unequal, the macro needs to print them and does so with the Debug trait.

You can take a look at the signature of assert_failed in core::panicking.

Approved some time ago

Approved

Approved

The point of this kata is not to write a function boolean isDiv5(String s), but to create a regular expression that will match all strings that represent in binary a number divisible by five and only these strings.

For example, this regular expression should match "101", "0" and "1111" because they represent 5, 0 and 15.
But it should not match "111", "10000" nor "1" as they represent 7, 16 and 1.

You mad genius

Approved

Approved

You are doing WAY too much work generating all the permutations. Let's take the example in the description :

The whole list of permutations of "abc" is ["abc", "acb", "bac", "bca", "cab", "cba"], and the answer is "bac".

Do you really need to know that "abc" comes before "acb" to determine the middle permutation ? No you don't.
Do you even need to generate them ? Once again, no you don't. After all, 'a' is the very first letter in order, so there's not much chance for it to be the middle permutation's first letter.

It's really quite simple, actually : look at the fork.

Edit : similar solution (except that I used 128 instead of 1000)