Yes you need to allow that

Prompts: aistudio.google.com/prompts/1uwR...

The example app given in the first prompt: github.com/multisynq/mu...

Thank you

I see we had the same idea. Nice!

In a way I like this best because it’s straightforward to explain:
* No damage? Nice, shield doesn’t break
* This attack breaks the shield? Okay, return the number of this attack
* Otherwise, this attack does damage, and we look at the next attack

I’ll admit that “straightforward” is … subjective 😇

Played a bit more – recursion is fun!

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codepen.io/codefrau/pen...

findShieldBreak = (dmgs, shld) =>
dmgs.findIndex(dmg => (shld -= dmg) < 0)

👸🏻🛡️🔥🔥🔥🐉

codepen.io/codefrau/pen...

Especially using single digit numbers to avoid having to tokenize was brilliant. It let us focus on the actual problem instead of incidental complexity

It’s not very readable, admittedly, even though it’s very succinct. But whittling it down to almost a one-liner was a fun exercise. Thank you for the prompt!

Okay now that I’ve seen all the other submissions I wonder if an actual interviewer would object to my use of `eval()`. Then again, the `splice(-2)` is neat to get the operands in the right order for sub/div. They should appreciate that

It’s been a while, yes, but I’m still reading your letter every week. One of the few highlights right now 💜

It’s how you combine subexpressions in RPN (see en.m.wikipedia.org/wiki/Reverse...)

Try this:

evaluatePostfix('123*+123++*')

Clever! But evaluatePostfix('123++') should be 6 🤓

That was fun!

function evaluatePostfix(s) {
let a = []
for (let c of s)
a.push(isNaN(+c) ? eval(a.splice(-2).join(c)) : c);
return a.pop()
}

My CodePen does the infix conversion first and evals at the end:

codepen.io/codefrau/pen...

When I work with JWTs this is incredibly helpful: jwt.io