 Goldman Sachs

## Questions d'entretien

Entretien pour Software Engineer Intern

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# 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?

## Réponses aux questions d'entretien

48 réponse(s)

171

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 le

18

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 le

14

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 le

15

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 le

7

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 le

3

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 le

1

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 le

1

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 le

2

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 le

1

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.

Utilisateur anonyme le

0

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 le

0

Utilisateur anonyme le

1

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 le

0

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 le

0

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

Mohinder le

0

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 le

0

TvRef

Utilisateur anonyme le

0

Tom le

0

n = 8 log2(n) - > 3

Utilisateur anonyme le

0

Accidentally posted before posting the explanation... - The definition of identical is being similar in every way. Therefore, the balls must weigh the same, or they are not identical at all, but simply similar.

D.F. le

0

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 le

2

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 le

0

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 le

0

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.

Utilisateur anonyme le

0

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

Md. Kauser Ahmmed le

0

No idea

Utilisateur anonyme le

0

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

K.White le

0

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

Utilisateur anonyme le

0

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.

0

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 le

0

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 le

1

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 le

0

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 le

0

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 le

0

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 le

0

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 le

8

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

VK le

1

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.

Utilisateur anonyme le

1

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 le

5

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 le

1

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 le

0

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. le

1

None. They are identical. None is heavier.

HSS le

3

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 le

0

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 le

5

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 le

3

None- weigh them in your hands.

Amy le

40

3 times. (2^3 = 8)

Brian le

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