modulo operation was expensive isn't it? it can be done without using module operation at all..

建议检查下长度
if str2.matches('0').count() == str2.len() {
return 1;
}

当str2为'0000'或者大于1个'0'的时候会出错

you can see his solution... there you will see why

impressive!

well it's an O(n) solution and there is an obvious O(1) solution

that's why it does not deserve its insane 250 upvotes

explain further since you aren't making any sense.

Awesome!

I'm only doing Rust Katas because I've never used it before. It's nice to say why you think iterators are no good here, otherwise you're sort of trolling.

Iterable solution makes no sense as best practice here

You are not right. The only information the range contains is start index and end index and no additional information is used to generate values from range, so this solution actually requires constant memory.