## Pregunta de entrevista

Entrevista de Software Engineer Intern

-New York, NY

Goldman Sachs## Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale . What's the fewest number of times you have to use the scale to find the heavier ball?

## Respuestas de entrevistas

47 respuestas

Two. Split into three groups of three, three, and two. weigh the two groups of three against each other. If equal, weigh the group of two to find the heavier. If one group of three is heavier pick two of the three and compare them to find the heaviest.

Marty en

2 3a+3b+2 = 8 if wt(3a)==wt(3b) then compare the remaining 2 to find the heaviest if wt(3a) !== wt(3b) then ignore group of 2 discard lighter group of 3 divide the remaining group of 3 into 2+1 weigh those 2 If == the remaing 1 is the heaviest if !== the heaviest will be on the scale

le-impresario en

2 weighings to find the slightly heavier ball. Step 1. compare 2 groups of three balls. Case 1. if they are both equal in weight, compare the last 2 balls - one will be heavier. case 2. If either group of 3 balls is heavier, take 2 balls from the heavier side. compare 1 ball against the 2nd from the heavy group result 1. if one ball is heavier than the other, you have found the slightly heavier ball. result 2. if both balls are equal weight, the 3rd ball is the slightly heavier ball. Easy Shmeezi

dutchmobil en

The answer is 2. 1) Divide the balls into 3 groups. 2 piles with 3 balls each, 1 pile with 2 balls. 2) Weigh the 2 piles of 3 balls. If both piles are the same weight, discard all 6 and weigh the last 2 to find the heavier one. 3) If 1 pile of 3 is heavier than the other, discard the lighter pile and the pile of 2 balls. Weigh 2 of the remaining 3 balls from the heavier pile. If both of the weighed balls are equal, the last ball is the heavier one.

Kevin en

Brian - this would be correct if you in fact were using a weighing scale, and not a balance scale. The ability to weigh one group against another with a balance scale allows Marty's answer to be a correct answer. Although - the question as worded provides a loophole. If it had been worded as "What's the fewest number of times you have to use the scale to CONSISTENTLY find the heavier ball", then Marty's answer would be the only correct answer. However, it is possible that you could get lucky and find the heavier ball in the first comparison. Therefore, the answer to the question as stated, is ONE.

Bill en

Respectfully, the folks who are answering "3" are mathematically modeling the nature of the balance incorrectly. Performing a measurement on a balance scale is not binary. It is trinary. Each measurement gives you one of three responses: The left is heavier, the right is heavier, or they are equal. So while you do need three binary bits to specify a number from one to eight, you need only two TRINARY-DIGITS Formally, you want the smallest value of n such that 3^n >= 8. The answer is 2. Note that you could add a ninth ball, and still, you'd only need to make two measurements. Of course, the smarty pants answer would be one. Just pick two balls at random and be lucky to have chosen the heavy one. But you're not guaranteed to be able to do it in just one measurement.

Morgan Creighton en

English isn't my mother tongue... What is a balance scale? If looking up on Google, I find some devices with two bowls on a bar bearing in the center. Hence, the answer is once (if I'm luck enough to select the heavier ball in teh first measurement) If a balance scale allows to measure only one ball at a time, then it would take two measurements, unless you'd have more information on the weight, which is not listed here, hence doesn't exist in the context of the question^.

Greg en

Since there is only one scale available to weigh. You first divide the balls in half. Weigh each group, take the heaviest group. This is using the scale twice so far. Now, divide the previous heaviest group into half, weigh both groups. Take the heaviest. Divide this last group and take the heaviest. This is the heaviest ball. We have used the scale 5 times.

Walter en

Without judging by hand: Put 4 balls on one side, and 4 on the other. Take the heavier group and divide again, put 2 balls on one side, and 2 on the other. Take the 2 that were heavier, and put one on each side. You've now found the heaviest ball. This is using the scale 3 times, and will always find the right ball.

Derek en

It's one. The fewest number of tries on using a balance scale would be one. If you put one ball on each side and one is heavier, you have the found the heavier ball.

Anónimo en

Use an equilateral triangular lamina which is of uniform mass throughout. It is balanced on a pole or a similar structure. Steps: Place 2 balls at each corner (total 6 balls) i. if the odd ball is one of those, one side will either go up or go down. Now repeat the process with one ball at each corner including the 2 unbalanced ones. ii. if balance is perfect, repeat the process with the remaining two balls and one of the already weighed balls.

Aruna Thapa en

test answer 2016-01-12 00:34:07 +0000

Anónimo en

You would not be able to find a ball heavier than the others. All eight balls are identical; therefore, they must all be the same weight.

Wesley en

Correct answer has already been posted. I just want to contribute some theoretical analysis. Given N balls, one of them is heavier. Finding out the ball requires log3(N) trit of information. (trit is the 3-base version of bit). Each weighing may give you one of the three outcomes: equal, left-heavier, right-heavier. So the amount of information given by each weighing is upper-bounded at 1 trit. Therefore, theoretical lower-bound for number of weighings in the worst case is log3(N), which is actually attainable. So 27 such balls need only 3 weighings and 243 balls need only 5 weighings, etc.

Marcus en

2 as many have indicated above. The 3 is the kneejerk reaction but 2 is the correct answer.

Mohinder en

Marty's answer is correct, but he does not explain why. The logic of the balance scale is three-valued: . Its most efficient use is the recursive application of the three-valued logic until there is only one item left. The integral ceiling of ln(x)/ln(3) thus gives the fewest number of times you have to use the balance scale to find the uniquely heaviest ball of x balls. Ceiling(ln(8)/ln(3)) = 2.

Carl Peter Klapper en

TvRef

Anónimo en

Reviewing the answers from over the years has been fun! Some time I would like to be the interviewer to ask these kinds of questions. In first looking at the question, I thought, "probably eliminate as many as possible with the 4 and 4" but then why would it he a thought experiment, less one in an interview, if it was so 2 dimensional? Whether or not getting to the best answer is much of the point, being reductionist by ignoring details, like the context of who is asking, lead me to go my own way as of the question was just text on a screen. There's a lot of 3 dimensional information you could get from someone by this question -- how nervous they are, if they don't then you see how they handle that, or how much they think of their answers to anything. I wonder, were the semantic holes in the question also intentional? In the comments people have written about technicalities -- things about how it wasn't specified that they were only visually identical, and therefore the question is contradictory. Or, how it didn't explicitly specify how to consistently, by most likely repeated efficient scenario, find the heavier ball, so people started that the least possible would be 1 if lucky. Then, since it wasn't explicit that you have to know which one was heavier, people said you could go 0 and guess the right one without the scale, either by direct guessing or trusting your hands with an unspecified sized difference in weight. If the word choice is designed to allow for those, perhaps the question is even more fun than I thought. One could see where and if one goes or gets hung up, see if one would ask clarification if they got hung up, or claim steadfastly about their thoughts being the most important on it, or focus elsewhere -- then that would be another layer to the question that makes it more interesting than I originally thought. I like tea.

Tom en

n = 8 log2(n) - > 3

Anónimo en

Fewest - get lucky and pick the heaviest one. But wait! How would you know it is the heaviest one by just weighing one ball? Your logic is flawed. Two groups of four. Split heavier one, weigh. Split heavier one, weigh. 3 times.

erich en

The answer is 2, as @Marty mentioned. cuz its the worst case scenario which u have to consider, otherwise as @woctaog mentioned it can be zero, u just got lucky picking the first ball....

whizkid en

Would it be wrong to say for a sample size as small as 8, we might as well not waste time thinking about an optimal solution and just use the scale 7 times, as this will be more efficient than coming up with an ideal solution prior to using the scale?

Craig en

I stumbled across this while looking for something else on Google but I had to answer. It is 2, split balls into 2,3 and 3. weigh the 2 groups of 3 against each other. If equal weigh the group of 2 and the heaviest is obvious. If they are not equal keep heavy group of 3 and weigh 2 of the balls. if equal heaviest ball is one you didn't weigh. If not equal the heavy ball is obvious.

Anónimo en

2 times. 8 balls. 1st step: [3] [3] [2] 2nd step: [[1] [1] [1]] [[1] [1] [1]] [[1] [1]]

Md. Kauser Ahmmed en

No idea

Anónimo en

The fewest number of times to use the scale to find the heavier would be Eight to One times ?

K.White en

It will actually be 1 because the question asks what's the fewest amount of times which is one because you could just get lucky you can use any method you want it would still be one because that is the fewest amount of turns you can have

Anónimo en

The point of these interview questions is to both check your logical brain function and to hear how you think. Most of you are just posting jerk off answers trying to be funny, or you are really dumb. These answer get you nowhere with me in an interview. Think out loud, go down the wrong path back track try another logic path, find the answer. None of this "0 if you use your hands". That is fine if you are interviewing for a job in advertising where creativity is desired, nobody wants you writing code like an 8 year old.

PuddinHead en

The problem is based on Binary Search. Split the balls into groups of 4 each. Choose the heavier group. Continue till you get the heavier ball. This can be done in log(8) (base 2) operations, that is, 3.

Vaibhav en

minimum is 1 (if lucky - 25% chance by picking 2 balls at random) & max is 2 (using most efficientl process to absolutely determine without luck - 3/3/2 scenario)

Brian en

1=if all the balls are identical and you pick up the first...balance it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.

Wanda en

Amy is 100% correct for the following reason: everyone (except Amy) is solving the theoretical problem. The practical side of the problem - notwithstanding jimwilliams57's brilliant observation that if one weighs more than the others IT IS NOT IDENTICAL (would have loved to see the interviewer's face on that one) - in order to 'weigh' them on a scale, one has to pick them up, therefore, you will immediately detect the heavier one without weighing: pick-up three and three ... no difference, no need to weight. Pick-up the remaining two to determine the heavier one. Steve

Steve en

First off, take yourself through the process visually and forget square roots, that doesnt apply here and here is why: The question is the Minimum, not the MAXIMUM. BTW, the max would be 7 ( 8-1); you are comparing 2 objects, so 1 ball is eliminated automatically in the first step. Anyway, you have a fulcrom of which you are placing 2 of 8 objects on each end. If by chance you pick the slightly heavier object as one of the two balls, you have in fact, found the slightly heavier one in the first round... btw dont be a smartass with your interviewer, he is looking for smarts not smarmy;)

John en

2=if all the balls are identical and you pick up the first...weigh it and the second one is lighter or heavier then you've found the heavier ball in the least amount of attempts.

Wanda en

While Symantec was busy weighing my balls I took a job with NetApp.... They need to focus on hiring good, capable security engineers, not weighing their balls.

Dash en

This question is from the book "How to move Mt Fuji".... Marty has already got the right answer.

VK en

It's 2. Period. If you can't figure it out look it up online or in "How Would You Move Mount Fuji" (like somebody else said). This is one of the most basic brainteasers you could be asked in an interview.

Anónimo en

You have 12 balls, equally big, equally heavy - except for one, which is a little heavier. How would you identify the heavier ball if you could use a pair of balance scales only twice?

chelsea en

Actually Bill, by your interpretation of the question the answer is zero, because you could just pick a ball at random. If you get lucky, then you've found the heaviest ball without using the scale at all, thus the least possible amount of times using the scale would be zero.

woctaog en

just once. Say you are lucky and pick the heavy ball. One use of the scale will reveal your lucky choice

fewest possible times using the scale en

so once, or the creative answer zero if you allow for weighing by hand

most answers seem to calculate the expected (average) number of times. en

With the systematic approach, the answer is 3. But, if you randomly choose 2 balls and weigh them, and by coincidence one of these two is the heavier ball, then the fewest number of times you'd have to use the scale is 1. Although the real question is: are the balls truly identical if one is heavier than the rest?

jimwilliams57 en

None. They are identical. None is heavier.

HSS en

i think its 3. i would take it like this OOOO OOOO then OO OO then OO problem solved. i do this everyday. bye. praise be to allah. thats it.

Nishat Chowdhury en

Assuming that the balls cannot be discerned by physical touch, the answer is 3. You first divide the balls in two groups of 4, weigh, and discard the lighter pile. You do the same with the 4 remaining, dividing into two groups of 2, weighing, and discarding the lighter pile. Then you weigh the two remaining balls, and the heavier one is evident.

Aly en

None- weigh them in your hands.

Amy en

3 times. (2^3 = 8)

Brian en