Interview Prep: Answer the guy above you
Thought about this idea from a similar thread just for shits and giggles.
First I'll post a interview question and the next person will answer it as best as possible (slick, sly, analytical, whatever you think will be appropriate to say that will help you get the job) and post a new interview question.
Stolen Example:
Interview Question: If you had the company's credit card for the weekend, what would you do with it?
Response: Depends on the limit.
Lets try it.
Interview Question: If you were shrunk and placed inside a blender, how would you escape?
*Don't forget to post another question after you answer!
I would stand in the middle so the blades wouldn't hit me or I would hide under the blades.
Tech Question: You have 12 balls and one is heavier or lighter than the others. Tell me which ball it is and if it is heavier or lighter in 3 weighings with a balance scale.
Since no one answered this question correctly I'll give it a shot. Assuming that you mean balance scale as a scale with two sides, this is what I think:
Break the 12 down into 3 groups of 4, and what you basically have to do is assign a unique sequence to each (without mirrors).
So:
So if the left side of the balance goes down on the first weigh in but is even for the other two you look for L - -. That means that ball #9 i sheavier than the others. If you have left side down on first weight in, then up on second weigh in, and back down on last weigh in you look for LRL, so ball #7 is heavier than the others.
I think that works.
As for an interview question I was once asked how to make an office building with 12 floors and 3 elevators more efficient (how to make the elevators not overcrowd and slow down).
EDIT: had to change this a few times since I kept messing up the sequence, but I think the one I have up now is correct.
you have the right idea--3 groups of 4 are what works, and you start off by weighing groups of 4 against each other and then breaking it down from there. the solution is in a lot of places online, as it's a pretty classic combinatorics question
on a related note, what's the relationship between the number of balls and the number of weighings necessary? i.e., come up with a formula relating n balls and w weighings if we have the same heavy/light scenario.
another classic problem: suppose I have a sequence of numbers 1,2,3.....100 (i.e. integers from 1 to 100) but I am missing one of the numbers in the sequence. What is the fastest way to determine which number I am missing?
here's one that you might really get. two bonds are trading at significantly different levels. same maturity. same coupon. why? give me a few potential reasons.
I would put all 12 balls on the scale and remove them one by one till there was a variance in the weights.
3.50x350÷√350
Case Study: Let's pretend I am blind, explain to me the color blue.
That is 12 weighings yourworstenemy.
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