# Brain teaser Interview Questions

interview questions shared by candidates

## Brain teaser Interview Questions

### Linux Kernel Engineering at Google was asked...

You have 25 horses, what is the minimum number of races you can find the top 3. In one race you can race 5 horses, and you don't have a timer. 29 AnswersI don't know, but Google it, you'll find the answer. Once you know it, it's obvious. 12 Probably 8. First answer: at least 5 (you have to look at every horse). Second answer: 11. You can race five horses, replace 4th and 5th place horse by two horses from pool of horses that haven't raced, and continue doing this until all horses have raced. Since there are 25 horses, you need 1 race for the first five horses and 10 to go through the remaining 20. Third answer: 8. Round 1 (5 heats) : Race horses in heats of five. Eliminate all 4th and 5th place horses. (15 horses left) Round 2 (1 heat): Race winning horses from every heat. Eliminate 2nd and third place horses that lost to horses in round 2 that came third or worse. (7 horses left). Round 3 (1 heat): Pick any five horses. Race them. Eliminate bottom two and replace with remaining two horses. (5 horses left) Round 4 (1 heat): Eliminate bottom two. Note: after round 2 it may be possible to use some interesting decision tree mechanism to determine the three fastest horses using fewer horses per heat but you still need two heats to do that. This problem assumes horses don't get tired, no ties are possible and a horse's speed is deterministic race after race after race. Good assumptions for brain teasers, but not for real life. Show More Responses Actually, should be 7. Forgot that after round 2 you can eliminate one more horse leaving you with 6 horses, one of which (the winner of the second round), is the fastest horse of them all. One more race with the five horses not No.1 and you pick the top two. 6. First round consists of 5 races of 5 horses. Pull aside the winner of each race in the first round for the final heat and eliminate the bottom 2. 8. Run 5 races of 5 horses. (5) Put all the 3rd place horses in a race, winner is 3rd fastest. (6) Put all the 2nd place horses in a race, winner is 2nd fastest. (7) Put all the first place horses in a race, winner is fastest. (8) The third place horse was beaten by one of the 2nd place horses. So third stays in third. Same for the second place horse. chzwiz, r u sure u r not high? B, what happened in round 2? why 8 horses suddenly get eliminated by 1 race? again, do u know what you are smoking? oh, forget J,..... just speachless 7. chzwiz, your answer is incorrect. One of the first or second place horses could be 2nd fastest and/or 3rd fastest. I drew a grid to visualize the problem. First, run five races to establish 5 groups of internally ranked horses, and you can of course immediately eliminate 4 & 5 of each race. 1 2 3 x x 1 2 3 x x 1 2 3 x x 1 2 3 x x Then race 1st place horses, eliminate 4 & 5, and those they beat earlier. You can also eliminate the horses #3 beat, and the 3rd place horse from #2's first race. 1 ? ? X X 2 ? X X X 3 X X X X X X X X X X X X X X You know #1 is fastest. Race the remaining horses (2, 3 and ?'s), and the top two are 2nd and 3rd. After reading the above answers, this is the same as B's revised answer, but I found it easier to explain/visualize with the grid. @brad ... but isnt dat v dont have timer to know d ranking ... or my assumption is wrong?? I thought v know only 1 result of every race .. i.e. top 1 Spawn, it's true that you don't know timings, but you do know RANKINGS. I don't see errors in logic with each answer above. They may claim to discover the answer with fewer races, but they have too many assumptions or errors to be correct. To do this optimally and without any potential error, you must eliminate as many correctly placed horses as possible and eliminate the maximum number of non-win/place/show horses each round. Round 1: (Races 1-5) You must start with 5 heats @ 5 horses apiece. The non-Win/Place/Show horses of each round have been eliminated, leaving 15 horses. Round 2: (Race 6) You then race the top finishing horses against one another, leaving you with your overall fastest horse. We now have the Win horse, leaving only Place and Show as unknowns. This also eliminates the bottom two horses, leaving 2 horses from this race. There are also only two more unknown horses. Race 7) You may race the Place horses against one another. Only two survive this race, since we know that each of these horses can place 2nd at best. This leaves 2 horses from this race. (Race 8) You may also race the show horses of each round. Since they each have been beaten by two other horses in Round 1, only the 1st place horse of this round survives as the potentially 3rd best horse. This will leave 1 horse in this race to battle for the 2 remaining spots. Round 2 leaves 2 Horses (Race 6) + 2 Horses (Race 7) + 1 Horse (Race 8) = 5 Horses, for 2 unknown spots. Round 3: (Race 9) We can now have the final 5 horses race for the remaining two spots leaving us with the undisputed 2nd and 3rd places. Here is a more complex scenario below in which the top 2 horses come from the same original race and the later races are full of lame horses. Many of the solutions above do not address those problems. Each # is a horse, named by it's inherent ranking. X signifies 'eliminated horse' Race1 R2 R3 R4 R5 1-->R6 3-->R6 11-->R6 16-->R6 21-->R6 2-->R7 7-->R7 12-->R7 17-->R7 22-->R7 4-->R8 8-->R8 13-->R8 18-->R8 23-->R8 5X 9X 14X 19X 24X 6X 10X 15X 20X 25X Round 2: Race 6 (Winners) Race 7 (Places) Race 8 1 -->1st Place 2-->Race 9 4-->Race 9 3 -->Race 9 7-->Race 9 8X 11-->Race 9 12X 13X 16X 17X 18X 21X 22X 23X Round 3: Race 9 2P 3S 4X 7X 11X First Place = 1 Horse -- Race 6 Winner Second Place = 2 Horse -- Race 9 Winner Third Place = 3 Horse -- Race 9 2nd place To fix the formatting issues that appeared as spaces are eliminated above-- Race1________R2___________R3___________R4___________R5____ _1-->R6_______3-->R6_______11-->R6_______16-->R6_______21-->R6 _2-->R7_______7-->R7_______12-->R7_______17-->R7_______22-->R7 _4-->R8_______8-->R8_______13-->R8_______18-->R8_______23-->R8 _5X__________9X___________14X__________19X___________24X _6X__________10X__________15X__________20X___________25X Round 2: Race 6 (Winners)__________Race 7 (Places)__________ Race 8 1 -->1st Place_____________2-->Race 9______________4-->Race 9 3 -->Race 9_______________7-->Race 9 _____________ 8X 11-->Race 9______________12X____________________13X 16X_____________________17X____________________18X 21X_____________________22X____________________23X Round 3: Race 9 2P 3S 4X 7X 11X Less ascetically pleasing than the original, but it will suffice. Show More Responses This problem can be reduced to (viewed as) multiway mergesort: 1) split the horses into groups of five as if we had five files to sort 2) do a multiway mergesort but instead of sorting all horses, stop as soon as you get the top 3 We need one race per group to sort them (5 races total). And 3 more races to get the top 3. We will need 8 races. The interviewer might have been looking whether an applicant was able to spot this relation to algorithms or not. But if this was a brain teaser instead, maybe the answer is less than 8. The answer is definitely 7, here is a fantastic explanation: http://www.programmerinterview.com/index.php/puzzles/25-horses-3-fastest-5-races-puzzle/ Randomly race the horses 5 times and keep a winner's bracket. The 6th race will determine the fastest horse. This is trivial. The 2nd and 3rd fastest could be in the winner's bracket or they could be the 2nd or even 3rd fastest in one of the losing brackets. Then we have to race 2nd and 3rd place from the winners bracket as well as from each original bracket, which had 5. This means we have 12 horses to try. Randomly keep 2 horses out and race 5v5. Keep the top two horses from both. We have 6 horses now. Race 3v3 and keep the top two, leaving 4 horses. Then race these. I count 11? Brute force divide and conquer solution = 18 races. Just be aware of the brute force solution in case there's restriction where we can't use the memory to save the first 3 places. Those who want a system that averages that fastest solving time, the answer will be 7 races.. Those gamblers who want an algorithm that gives the fastest possible solving time (all though sometimes it WILL take longer) the answer is 6. Race 1-5: First 5 races of five.. We randomly seed with no overlaps.. we use this to eliminate the bottom two from each race. I am going to use a two digit number labeling system where the first digit is which race in heat one they competed in and the second digit is which place they obtained in that heat.. These are the horses we have left: 11 12 13 21 22 23 31 32 33 41 42 43 51 52 53 Race 6: We have the winners from each heat compete, and for ease of explanation lets say that they ranked 11, 21, 31, 41, 51.. Now we know that 41, 51 and all the horses slower than them (their heats) are out.. also, we know that any horse slower than 31 is out (ie. 32 & 33), and we know that any horse that is at least two horses slower than 21 is out (ie 23). This leaves 11, 12, 13, 21, 22, 31.. 6 horses. Possible rankings: 11 12 13 21 22 23 31 32 33 Race 7 Rank 1 is already known.. and we just need to know how to organize 12, 13, 21, 22, and 31.. And look at that!! that's 5 horses! We race those 5 and the top two receive the overall rank of 2nd and 3rd. 5 races in round one, 1 race in round 2, 1 race in round 3.. it took us 7 races.. Option two: 6 Races Now, an alternate answer.. when we read "What is the minimum number of races to find the top 3 horses" we can interpret that maybe they don't care about this method working every time.. instead they want to hit up the Vegas Casinos and test their luck, and maybe save some track time.. If i was a gambling man.. i would race the first group of those that i thought looked the strongest.. I would eliminate 2 in that race (we are now down to 23 horses).. and then race the 3rd fastest multiple times and hope that he gets first in the remaining races, thus eliminating 4 horses each time. This is going to be painful for the horse that placed 3rd in heat one. Race 1: Eliminate 2, 23 horses remaining Race 2.. Eliminate 4, 19 horses remaining Race 3.. 15 horses Race 4.. Eliminate 4, 11 horses Race 5.. Eliminate 4, 7 horses Race 6.. Eliminate 4, 3 horses.. The original top 3 from Race 1. So using method 2, we completed this race in 6 races instead of 7! I messed up the possible ranks after race 6.. That should have read.. 11 12 13 11 12 21 11 21 22 11 21 31 Please forgive me. Brad's explanation is the perfect one. All other explanations are junk and impossible to read and follow. Round1: (5 groups) remaining horse 15 Round: (1 groups). Topper from each heat Round: (1 gounds). Forget about the 4th and 5th Heat horses (becoz their toppers are at last place) go to the 3rd position 6 races as the 25 horses are divided into 5 groups each grouping 5 horses in each group , then the winner from each group would be 5 then there should be last n final race to find the first , second and third placed horse . in this way we can find our top 3 horses :) It should be 11. In each race you can only eliminate 4th and 5th place simply because there's a chance that the top 3 in that single heat are the 3 fastest of all the horses. So each race you can select 5 random horses each race or keep the top three in the following race, the outcome is the same. Show More Responses I get 8 races to guarantee you get the top three. 5 races of 5 to get the fastest horse from each group. 1 race to determine the fastest overall. The second fastest horse can be in the group (x) that produced the fastest, 1 race with the remaining 4 fastest from the other groups and the second fastest from group x. That race will produce the second fastest overall guaranteed. 1 more race with the 3 or 4 from the original winners and the replacement from the group that produced the second place horse. That makes 8 races. I dont get it.. It shoudl be five races right? U will know how long each horse takes to cover same distance. Pick the three that took shortest. Y need more rounds? If u down vote my answer, u have the obligation to explain why 😜 Answer is 6. You will have 5 races and 5 winners. Next race with 5 winners is the 6th race and you get top 3. You're welcome. It would be 7, mainly because since there only 5 racers each round, then it would be tournament style. As you can tell, tournament style never determines the rankings in the short term. (Imagine if a new tennis player played against Roger Federer and he loses. He's not bad, he's just really unlucky haha) So in order to determine the answer, you would have to use massive logic and reasoning. First step is obvious: you get 5 sets of horses to race through 5 races. So the number is already at 5. We get the top horse in each one and note them. However, one problem is that you have to consider if fastest horses could lie all in the first, second, third, fourth, or fifth group. But we can't determine that yet, so let's race one more time with the winners of each race. Second step: Race with each winner from their groups. That makes 6 total races so far. Ok, now we can get somewhere with this. So now, its a question of which horses we can eliminate to determine the next race. Who can we eliminate though? Well, we can get rid of: - All the horses in the group with the last race that were 4th and 5th place. This makes sense since if the winner of their groups only made 4th and 5th place in the tournament style race, then they are definitely not the fastest horses. 15 horses remaining. - Well, we can also get rid of the last two horses of each group since we are only looking for the top 3. 9 horses remaining. - We also can get rid of the first place winner since we already know he is definitely the best. 8 horses remaining. - Since we already determined who the fastest horse is, now we have to determine who the second and third fastest horse can be. In that case, the second place horse from the last round can get rid of the third horse of that group, since he definitely has no shot of being the first or second fastest horse of this new race. Also, in the third group, we can get rid of the second and third horse in the group due to the same reasoning above. 5 horses remaining. Oh wow! We now have 5 horses left! This means we can just have a clean race and find the second and third fastest horse. So overall, the answer is 7 races! 7 In my opinion, you need 26 RACES to make sure that you get the top 3 without a timer. You need to know that without a timer if we plan only 5 races with 5 horses on each one, the last horse on the first group could end the race in 1 minute while the first horse on the second group (the winner) could end it in 2 minutes, which mean that the last horse of the first group is much faster than the first horse of the second group..and this is why we need to make a total of 26 RACES to determine the top 3 as follow: 5 RACES with 5 horses each .> We get the 5 winners from each race .> We race the 5 winners > We get the top 4 > We race the top 4 with one of the remaining 20 horses > we get the top 4 > we race the top 4 with one of the remaining 19 horses > we get the top 4 > we race the top 4 with one of the remaining 18 horses...etc until the last race then we select the top 3 winners from the last race. 5 RACES + 1 RACE + 20 RACES = 26 RACES |

How many passengers leave JFK airport on a given day. 19 AnswersNot sure: approx. 100,000? 1.5 million no matter if they are flying in or out, they all leave Show More Responses My answer would have been the same: "Pretty much all of them!" :) planes leave, people board. ALL OF EM ! (and to elaborate - The no. of ppl flying in + the no. of ppl flying out --all leave basically some on foot and other on air All leave except for Tom Hanks, he stays stuck in "The Terminal"....... (You have to include some humor when you answer :) 50k how is this a brain teaser? This question tests how you think critically. In this example, you could start with estimating that JFK has about 100 gates, and that a plane leaves from a gate every 3 hours, totaling 800 flights per day. Guessing that the average plane holds 200 passengers, you arrive at 160,000 per day departing. The interviewer is not as interested that you get the right, or even close, answer, but rather the thought process you go through to figure it out. For those that are interested, the real number is about 131,000 per day based on statistics from 2007. the question was "How many passengers leave JFK airport on a given day.?" my answer is the (number of people arrives that day )+( the employees work there (except for the security guards)) after all no-one stays on the airport forever The answer is the thought process, as above. Personally, I'd have started with the number of flights: flight frequency per runway * runways * open hours. Assume half are leaving, then guess the number of passengers per plane. I might also point out that the number of assumptions multiplied together in that calc is such that you're unlikely to be within an order of magnitude of the right answer... Tim Besse stole my answer! And I thought I was the only one who thought of that. Seriously though, if I was asked this question in an interview I would answer "probably about half" figuring that half the flights are departing and half are arrivals, similar to what another poster stated. Depending on the position you are interviewing for they may be impressed by you going through a long critical calculation, I don't know. In my line of work, character is more important that pure intelligence. Show More Responses all of them 100% 1. Those passengers that arrive on an airplane either leave by another plane or leave the airport by other means. 2. Those passengers that arrive by other means take a plane and leave or they are not passengers. 3. That is, if you exclude those that are still there at midnight. You can assume this number is constant from night to night. Well, there are passengers in cars that come and go, those onboard planes that do the same and some perhaps catching a lift on a bicycle or maybe even piggyback. If there are those that try to actually ride on JFK, he's dead, so - good luck. I would have to agree with those that say all of them. However I must point out an addition to the facts. We are all assuming something when answering these types of problems and as such the answers are as much subjective as they maybe factual. To the rest here, READ the question. Passengers are ON the planes if not then they are NOT passengers! ALL passengers leave eventually however what is a 'day' not specified as 24hrs- because a 24hr period COULD span 2 'days'. ALSO it is possible that a plane could be held hostage on the tarmac for more than 24hrs - these passengers would not leave on the same 'day' so our total may be slightly less than 100% on any given day. Ultimately though, I agree that it is the interviewer's 'hidden' question that counts. With these types of questions the answers are not so mathematically obvious. But rather, the questioner is looking deeper into your logic, character, personality etc. Are you a thinker, do you think out-side the box, do you get frustrated, are you a smart-ass, etc. That's been my experience not all them. because som arrive at one day and leave the next day. My answer will be "A LOT!" the answer does not require a number but how will you find out. It depends on the day of the year and also on how many people are actually on the plane and if all of them are going to JFK or are catching connecting flights. |

### Software QA Engineer at Apple was asked...

There are three boxes, one contains only apples, one contains only oranges, and one contains both apples and oranges. The boxes have been incorrectly labeled such that no label identifies the actual contents of the box it labels. Opening just one box, and without looking in the box, you take out one piece of fruit. By looking at the fruit, how can you immediately label all of the boxes correctly? 38 AnswersAll the three boxes are names incorrectly. SO the bax lebeled Apples+Oranges contains only Oranges or Only Apples. Pick one fruit from it. If it is Orange then lebel the box as Orange. So the box lebeled Oranges contains Apples and the remaining contains both. Label the boxes fruit. The key bit is "All the three boxes are names incorrectly" so the label on the box which fruit comes from will need to be changes to one of the other 2 labels. It can only be 1 of them (and it will be obvious when you have the fruit) then the remaining box (that hasnt featured yet)...Just swap that label with fruit box that was originally on the box which you took the fruit out of Thats hard for anybody to understand somebody elses explanation... eaiest way is to just do an example Show More Responses Swaz answer is almost correct however it does not work in all scenarios. lets assume: box 1 is labelled Oranges (O) box 2 is labelled Apples (A) box 3 is labelled Apples and Oranges (A+O) and that ALL THREE BOXES ARE LABELLED INCORRECTLY" Pick a fruit from box 1, 1) if you pick an Orange: - box 1's real label can only be O or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - box 1's new label should then be A+O by elimination - since ALL LABELS ARE INCORRECT - box 2's label is changed to O - box 3's label is changed to A - SOLVED 2) if you pick an Apple: - box 1's real label can only be A or A+O - box 1's current label is O - since ALL LABELS ARE INCORRECT then box 1's real label can not be O - this still leaves us with the choice between label A and label A+O - which would both be correct - FAILURE Solution: The trick is to actually pick a fruit from the A+O labeled box Pick a fruit from box 3: 1) if you pick an Orange: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be O by elimination - since ALL LABELS ARE INCORRECT - box 1's label is changed to A - box 2's label is changed to A+O - SOLVED 2) if you pick an Apple: - box 3's real label can only be O or A - box 3's current label is A+O - since ALL LABELS ARE INCORRECT then box 3's real label can not be A+O - box 3's new label should then be A by elimination (not O) - since ALL LABELS ARE INCORRECT - box 1's label is changed to A+O - box 2's label is changed to O - SOLVED it only says you can't look, doesn't mean you can't feel around or smell the fruit you picked, easy deduction after you figure the first box out Sagmi is right, but did not give the full reasoning. "the bax lebeled Apples+Oranges contains only Oranges or Only Apples. Pick one fruit from it. If it is Orange then lebel the box as Orange." so far so good Now, the box labelled Apples cannot be the box containing only Oranges, you've just found that box, so it must contain Apples and Oranges. And in that case the other box, labelled Oranges, must contain only Apples. It's easier to draw it out. There are only 2 possible combinations when all labels are tagged incorrectly. All you need to do is pick one fruit from the one marked "Apples + Oranges". If it's Apple, then change "Apple + Orange" to "Apple" The "Apple" one change to "Orange" The "Orange one change to "Apple + Orange" If it's Orange, then change "Apple + Orange" to "Orange" The "Apple" one change to "Apple + Orange" The "Orange" one change to ""Apple" Since all 3 boxes are labled incorectly Start with the box Labled A&O. If Its apples than the box labled apples then the apple one is oranges and the oranges is O&A. Label each box "Apples and/or Oranges" and the all will be correct. This is very simple to resolve. I was asked the same question at FileMaker. Each box is incorrectly labeled. So you go to the box that is labeled "Oranges and Apples" and take one out. It doesn't matter what comes out because all that you know is that it is not AO. If you remove an Apple then move the Apple label to it. Since the Apples are already identified it is easy to resolve the rest. All you know for certain is that the other two boxes remaining are mislabeled. So the AO label goes on the box with the remaining label and that label goes on the Apple box as you have already assigned that. The end result is you only need to remove one piece of fruit to figure out the proper locations of all. Go down the road to HP. Maybe they are hiring. Some of these pseudo-problem solving questions like this are bunk. I was once asked why sewer covers are round and not square. I gave the correct answer without even hesitation and the interviewer seemed put off that I knew the answer. I didn't get the job but, in hindsight, no great loss. I prefer the questions (like the basketballs one from google) where you won't be able to give an accurate numerical answer but by explaining HOW you would go about solving the problem is all you need to do and MAYBE shows your aptitude for problem solving. Smell the box you opened. Step 1: Order the boxes by weight. Either apples weigh more than oranges, or oranges weigh more than apples. The mixed box will always be in the middle. Step 2: Open the first box, take out the fruit and look at it. Step 3: If the fruit is an apple, deduce that the middle is apple and oranges and that the third box is oranges. If the fruit is an orange, then deduce that the last is the box with the apples. Show More Responses Donna is the only one with any common sense. The problem with corporate America, is that it's run by a bunch of Bozos who over complicate things and have a narrow to zero vision on how to solve even the simplest problems. I can imagine that most of you would get a committee, have long meetings where you talk about 'think out of the box', and 'at the end of the day' nonsense. This is an interesting logic question, but I would not want to buy fruit from a company who knew they had a problem and then sampled one out of three boxes to resolve the issue. There are other correct answers posted. I'll just make a comment: "The boxes have been incorrectly labeled such that no label identifies the actual contents of the box it labels." Nothing in the above statement says the labels are limited to oranges/pears, only that they do not identify the contents. They could say 'nuts', 'bolts', etc. Technically, all answers should be prefaced with: assume that the labels say 'oranges', 'pears', and 'orange/pears'. Ok, the problem does not make sense and is unsolvable if the labels say 'x', 'y', 'z', but someone with (likely with a math proof back ground) may appreciate attention to detail. Q: Why do posters denigrate the interview questions? The questions, however stupid they may be, are a opportunity to show you can build an answer. Even if you pursue an invalid train of thought in the interview, it's a thought. It's what they want to see and what will help you get a job offer. Note: I also would not assume that the questions asked are a reflection on the company, department, or team as a whole. It may just be the interviewer that has chosen poorly. So to say "I don't want to work for company X because they asked me a stupid interview question" is pretty closed minded. To even think I don't want to work with that interviewer just based on questions asked seems extreme. rightly pointed out by Sagmi ... this question was put forward to me at Huawei Technologies and that was the answer I gave So the question was asked at an interview for Apple: Label ALL the boxes apple and charge a ridiculous price for them! Just label all of them "Fruit." Put another way, it is not possible to tell since we don't know how the boxes are mis-labled. What if the Apple box was labeled Oranges and both the other boxes were labeled Apples and no box was labled Apples and Oranges? You might have assumed there are three different labels when their might have only been two different labels. Always pict a piece of fruit from the box labelled Apple&Orange. As we know that this label is wrong, there are two possibilities: If it is apple, then wo know that this box should be labelled Apple, so we switch Apple label with the label Apple&Orange. Then Apple label is correct. We also know that the Orange label is incorrect, so we then switch Orange label and Apple&Orange label. if it is orange, then we know that this box should be labelled Orange, so we switch Orange label with label Apple&Orange. Then Orange label is correct. The same as above, we know that the Apple label is incorrect, so we switch Apple label and Apple&Orange label. If all boxes are labeled incorrectly and u pick a orange out of a box that's labeled apple/oranges change the name to oranges then change the box labeled oranges to apples and the the box labeled apples to apple and oranges... If you pick a apple out of a box labeled apples and oranges change the name to apples and then change the box labeled apples to oranges and the last to apples and oranges... If u pick a apple out of a box labeled oranges change it to apples and oranges then the box labeled oranges to apples and the box labeled apples to oranges...if you pic a orange out of a box labeled apples change it to apples and oranges and the box labeled oranges to apples and the last to apples and oranges... See the pattern? I think there is a big box and it contain two small boxes and all the labels are incorrect so big one contain two boxes that makes it carrys both orange and apples and in that thwo boxes having orange and apple respectively so if we open any box we can label it correctly Show More Responses To see the java source code of puzzle, visit: https://github.com/SanjayMadnani/com.opteamix.microthon code is taking the input by console only. You can fork or clone the repository and proceed further. You can also rise bug if you find any. Run BasketPuzzleGameTest.java class as a Junit test case to start game. if it known already that boxes labeled incorrectly, I would give it back to those who did label them and ask to fix this confusion. it is impossible to tell by opening only one box, so you have to open one more box. As mentioned already, if you start with the A+O bucket, you can solve the puzzle by pulling only one fruit, Bucket: A+O Found: A Bucket A+O > A | A+O, but since A+O label is incorrect, then it must be A Bucket O > since A is taken, the new label must be O | A+O, but since O is incorrect, it must be A+O Bucket A > since A and A+O are taken, it must be O Bucket: A+O Found: O Bucket A+O > O | A+O, but since A+O label is incorrect, then it must be O Bucket A > since O is taken, the new label must be A | A+O, but since A is incorrect it must be A+O Bucket O > since O and A+O are taken, it must be A If you are lucky, you might solve it with just one fruit even if you start with other buckets, Bucket: A Found: A Bucket: A > A | A+O, but since the A label is incorrect, it must be A+O Bucket: O > A | O, but since the O label is incorrect, it must be A Bucket: A+O > since A+O and A are taken, it must be O Bucket: O Found: O Bucket: O > O | A+O, but since the O label is incorrect, it must be A+O Bucket: A > A | O, but since the A label is incorrect, it must be O Bucket: A+O > since A+O and O are taken, it must be A If you start with the A bucket and pull an O or if you start with the O bucket and pull and A, then you are SOL and you need to pull out more fruits to figure it out. 1. Open one box and check its contents. 2. Remove the current label and apply the correct one (by removing it from one of the other boxes) 3. Since all boxes have been labeled incorrectly, switch labels between the other 2 boxes. And Voila you have all the boxes labelled correctly :) In requirement already specify that all three box labels are not correct. A+O A O Step1: First Pick an item A+O Box. If you get an Apple then it is a Apple Box. swap the label . A A+O O AS we already know in the box that label with Orange, does not contain Orange because of wrong label. So It must contain A+O. Just Swap the label A O A+O OK, all 3 boxes are incorrectly labeled. Open the one that says apples and oranges. Whatever is in there is what it is (since it cannot be apples AND oranges). Now, if there was an orange in there, apples must be in the orange box (since they cannot be in the apples box), and apples and oranges in the apples box (due to process of elimination). Get it? I guess questions like these will appear easy if you put them on paper, it is the possible combinations that become relevant, one way to approach is.. One of the key factor is all boxes are labelled incorrectly, this gives rise to only (2) combinations right To label for 1st box incorrectly you will have (2) options, once you label it then you have only choice to label the other two boxes incorrectly so 2 x 1 = 2 combinations possible i.e. Incorrect lablling options { Boxes_with_Oranges, Boxes_with_Apples, Boxes_with_Apples&Oranges } = { A, AO, O} or {AO, O, A} 2. To know for sure the contents of the boxes, you need to pick the box with either Apples or Oranges and avoid box with Apples and Oranges. So from the (2) combinations you could pick a fruit from Box_labeled AO (this will contain either Oranges or Apples) So, if you get a Orange, it means that combination is{AO, O, A} , so that means Box_with_Label_O has Apples, Box_with_Label_A has Apples and Oranges Box_with_label_AO has Oranges or else if you get a Apple that combination is {A, AO, O}. Box_with_Label_AO has Apples, Box_with_Label_O has Apples and Oranges Box_with_label_A has Oranges Then you can correctly label all the 3 boxes. First answer in this post is correct, as its said all boxes doesnt reflect correct items in it, If an apple is picked from a box , then it can be from either A/O box or A box, if the box is names A/O the, the label of the box has to be changed to A, then other two box labels to be accordingly. It is interesting that in 6 years people keep overthinking this. The answer is in the question and the criteria are that the boxes are immediately labeled and they are labeled correctly. ANSWER: FRUIT FYI you don't even need to open one box. Show More Responses Your choice going to be (( 2 apple 1 orange)) or (( 2orange 1apple )) . It can be recognize only one box (x) . U have to chose again until u get another formula then u will named easly . Step 1: Open a box labeled ‘Apples and Oranges’. We know that this box does not contain ‘Mixture’ for sure. If this fruit is an apple, then label this box as ‘Apple’. Step 2: (Very important) If we look at the box labeled as ‘Oranges’, we know that since the label is incorrect, this box either has only apples in it or has Mixture. Since we already know which box contains only apples, we know that the box labeled as ‘Oranges’ contains ‘Mixture’. So label it as ‘Mixture’. Step 3: (Very easy) The 3rd box will be labeled as ‘Oranges’. When you put your hands on the box to pick the fruits by touching every fruits you can feel whether all are apple or oranges or both and just pick one to see.So it is not necessary to pick one fruit and see whether it is orange or apple also it is not said in question that you can touch and feel all the fruits inside the boxes without taking it out .and then you can fix the label correctly on the boxes. Absurd, no logic km I will took a pen and stab the apple. and then ? apple-pen. |

### Data Scientist at Facebook was asked...

You're about to get on a plane to Seattle. You want to know if you should bring an umbrella. You call 3 random friends of yours who live there and ask each independently if it's raining. Each of your friends has a 2/3 chance of telling you the truth and a 1/3 chance of messing with you by lying. All 3 friends tell you that "Yes" it is raining. What is the probability that it's actually raining in Seattle? 31 AnswersBayesian stats: you should estimate the prior probability that it's raining on any given day in Seattle. If you mention this or ask the interviewer will tell you to use 25%. Then it's straight-forward: P(raining | Yes,Yes,Yes) = Prior(raining) * P(Yes,Yes,Yes | raining) / P(Yes, Yes, Yes) P(Yes,Yes,Yes) = P(raining) * P(Yes,Yes,Yes | raining) + P(not-raining) * P(Yes,Yes,Yes | not-raining) = 0.25*(2/3)^3 + 0.75*(1/3)^3 = 0.25*(8/27) + 0.75*(1/27) P(raining | Yes,Yes,Yes) = 0.25*(8/27) / ( 0.25*8/27 + 0.75*1/27 ) **Bonus points if you notice that you don't need a calculator since all the 27's cancel out and you can multiply top and bottom by 4. P(training | Yes,Yes,Yes) = 8 / ( 8 + 3 ) = 8/11 But honestly, you're going to Seattle, so the answer should always be: "YES, I'm bringing an umbrella!" (yeah yeah, unless your friends mess with you ALL the time ;) I thought about this a little differently from a non-bayes perspective. It's raining if any ONE of the friends is telling the truth, because if they are telling the truth then it is raining. If all of them are lieing, then it isn't raining because they told you that it was raining. So what you want is the probability that any one person is telling the truth. Which is simply 1-Pr(all lie) = 26/27 Anyone let me know if I'm wrong here! Here's another perspective on how to answer a question like this: Bring an umbrella. It's Seattle - if it's not raining right now, it probably will be by the time you get there. Show More Responses I flagged Nub data scientist's answer as useful, because it shows an interesting flaw in reasoning. The 3 random variables are not to be treated as intrinsically independent. Only conditioned on the truth (raining/not raining) are they independent. Isn't the answer 2/3. The key thing is that they are ALL saying "Yes". You can't have all 3 says yes and have some people lying and some people telling the truth. It either is raining or it isn't. Not both. They either are all lying or all telling the truth. Since they are all in agreement (all lying or all truthful), they are essentially voting as one person. What is the probability that one person is telling the truth? 2/3 Answer from a frequentist perspective: Suppose there was one person. P(YES|raining) is twice (2/3 / 1/3) as likely as P(LIE|notraining), so the P(raining) is 2/3. If instead n people all say YES, then they are either all telling the truth, or all lying. The outcome that they are all telling the truth is (2/3)^n / (1/3)^n = 2^n as likely as the outcome that they are not. Thus P(ALL YES | raining) = 2^n / (2^n + 1) = 8/9 for n=3 Notice that this corresponds exactly the bayesian answer when prior(raining) = 1/2. I'm not sure why it's not just as simple as this: All three friends say it is raining. Each friend has prob. 1/3 of lying. Since the friends all say the same thing, they are either all telling the truth or all lying. The question asks what is the probability that it is raining. This is equivalent to asking, what is the probability that all three friends are telling the truth. And that is equivalent to asking, what is the probability that not one of them is lying. Since the the friends were asked independently, this should equal 1 - (1/3 * 1/3 * 1/3) = 0.962. Ah. Looks like my answer agrees with "nub data scientist". What is the probability that both he and I are wrong? :-) TLP and nub data scientists, Your answers include possibilities which are not feasible; we cannot have any combination of 2/3 and 1/3 together... what about (2/3)^3? I agree with TLP and nub scientist. For me, the question is really (1 - the odds that all three of your friends are lying to you) Clearly 1 - 1/3 * 1/3 * 1/3. It's convenient that they all gave the same answer, otherwise it would be more difficult. Let Y denote rain, N denote no rain Actual Answer probability ------------------------------------------ Y=> 8/27 YYY, 1/27 NNN, 12/27 YYN, 6/27 YNN N=> 1/27 YYY, 8/27 NNN, 6/27 YYN, 12/27 YNN So, P(Y|YYY) = (8/8+1) = 8/9 The probability of raining is that they are all telling the truth, therefore, (2/3)^3. P(rain / yes yes yes) = (2/3)^3 / ((2/3)^3 + (1/3)^3) =(8/27) / ((8/27) + (1/27)) = 8 / (8 +1) = 8/9 26/27 is incorrect. That is the number of times that at least one friend would tell you the truth (i.e., 1 - probability that would all lie: 1/27). What you have to figure out is the odds it raining | (i.e., given) all 3 friends told you the same thing. Because they all say the same thing, they must all either be lying or they must all be telling the truth. What are the odds that would all lie and all tell the truth? In 1/27 times, they would the all lie and and in 8/27 times they would all tell the truth. So there are 9 ways in which all your friends would tell you the same thing. And in 8 of them (8 out of 9) they would be telling you the truth. Show More Responses There is an obvious conceptual reason as to why several answers here (ones that don't use Bayes' formula) are incorrect. The probability in question has to depend on the probability of rain in Seattle. If, for the sake of discussion, it ALWAYS rains in Seattle, i.e. P(rain)=1, then the required prob. is always 1 as well. Likewise if it's a place where it never rains, or if the question asks about the prob. of it raining elephants given the 3 friends said yes, it'd be still 0. I believe this is a std. textbook example of the Bayes' formula, anything short of that I don't think will work out. Please correct me if incorrect. But I would just prefer to condition. either they are all telling the truth and its it raining or they are all lying and it is not raining. P(rain)=P(rain|truth,truth,truth)*P(truth,truth, truth)+P(rain|lie,lie,lie)*P(lie,lie,lie) notice that truth does not mean yes it is raining, it simply corresponds to them telling the truth. Since they said yes, IF they were lying and we knew they were lying then the probability of rain would be zero, thus eliminating the second term. P(rain)=P(rain|3xtruth)*P(3xtruth) and the probability of the truth is (2/3)^3 and the probability of rain if they are telling the truth is 1. I did a little skipping of steps, since truth doesnt equal yes, but i just sort of meshed it toegher towards the end YES=yes,yes,yes T=truth, truth, truth L=lie,lie,lie P(Rain|YES)=P(Rain|YES,T)*P(T)+P(Rain|YES,L)*P(L) P(Rain|YES,L)=0==> whats the probability of rain given we know that they are lying and theyve told us it is raining. P(Rain|YES)=P(Rain|YES,T)*P(T) P(Rain|YES,T)=1==> whats the probability of it raining given that they are telling the truth and have told us its raining then P(T)=(2/3)^3 its obvious. why in the world would i do bayesian methods when its certain I think the first answer is incorrect. The basic flaw is that it is assumed that all three friends lie together or be honest together, so it does not take the cases of Yes.no.Yes or Yes.Yes.no ...etc For the correct answer we need to update posterior probability after each yes so Assuming P(raining) =0.75 prior probabilty P(raining | yes) = (2/3)*0.75 / ( (2/3)*0.75 + (1/3)*0.25 ) = 6/7 P(raining | yes,yes) = (6/7)*(2/3) / ( 6/7*2/3 + 1/7*1/3) = 12/13 P(raining | yes,yes,yes) = (12/13)*(2/3) / ( 12/13*2/3 + 1/13*1/3) = 24/25 I dont see the interview saying that all friends are sitting together so they are independent which means they can lie separately I agree with (2/3)^3. Interview Candidate solves this problem using Bayesian stats despite the fact that no enough information is given to do Bayesian probability analysis i.e. he had to pull the probability of it raining in Seattle out of thin air when it was not given in the interview question. With only the information from the interview question, we have to assume that friends are either all lying or all telling the truth. Let truth=T and lie=L P(TTT)=8/27, P(LLL)=1/27, P(TLL)=2/27,P(TTL)=4/27. But we know that they all had the same answer, so we must compare P(TTT) to P(LLL). P(TTT) is 8 times more likely than P(LLL), so we have P(All same answers|TTT)=8/9, P(All same answers|LLL)=1/9. Therefore the solution given ONLY THE INFORMATION GIVEN is P(Rain)=8/9, P(Dry)=1/9. This problem requires the marginal probability of rain to solve, following Interview Candidate's answer. M.B. provides the rationale behind why the bayes approach is necessary: if the pr(rain) = 0, then the pr(rain|y, y, y) = 0. (maybe it is July in Seattle). A few conceptual problems in many answers that I want to point out: 1) There is lots of conflation between Pr(truth) and Pr(Y). Pr(truth) = Pr(Y|R) does not equal Pr(Y). 2) Consider there is only a single friend and they say yes, the logical conclusion from a lot of these answers is that Pr(Rain|Yes) = Pr(Yes|Rain) = 2/3, which is not correct. Bayes' rule is very clear in this simpler case. 3) The friends' answers are conditionally independent assuming no collusion. The combinations of their honesty/lying adds no additional information. The marginal probabilities are not independent, Pr(y,y,y) does not equal pr(y)^3, it equals pr(y,y,y,rain) + pr(y,y,y, no rain), the integration of the joint space over rain. Using conditional independence and bayes rule, this becomes: pr(y|rain)^3*pr(rain) + pr(y|no rain)^3(1-pr(rain)). A more general solution using Pr(rain) = r. Pr(rain|y,y,y) = Pr(y,y,y|rain)*pr(rain)/pr(y,y,y) #Bayes' formula pr(y,y,y|rain) = pr(y|rain)^3 = (2/3)^3 #conditional independence pr(y,y,y) = pr(y|rain)^3*pr(rain) + pr(y|no rain)^3*pr(no rain) #by definition, see point 3 the answer: r*(2/3)^3 / [r*(2/3)^3 + (1 - r)*(1/3)^3] It should be (2/3)^3, I think zen and todo is correct. As a big dumb animal, I have to write out a probability tree and thing about this simply. You only have 2 scenarios where all three say it is raining (all three are telling the truth-raining OR all three are lying - not raining). Assume the probability of rain is 0.5 for simplicity. P(Rain and YYY) = 1/2 * 2/3 * 2/3 * 2/3 = 8/54 P(Not Rain and YYY) = 1/2 * 1/3 * 1/3 * 1/3 = 1/54 Thus P(Rain | YYY) = P(Rain and YYY) / [P(Rain and YYY) + P(Not Rain and YYY)] = 8 / (8+1) = 8/9 I know it isn't the most mathematically rigorous or syntactically correct solution, but I'd bet a pretty penny that the answer is 8/9 with the following assumptions (P(rain) = 0.5 and naive bayes - friends didn't collaborate). Most of the answers/comments made all unconditional assumptions except a few reasonings that lead to the 8/9 probability. Note that the question states that "Each of your friends has a 2/3 chance of telling you the truth". This essentially means P(raining, yes) + P (non-raining, no) = 2/3. Any attempts to interpret this as conditional probability P(raining | yes) = 2/3 or P(yes | raining) = 2/3 are making other assumptions. Show More Responses 8/27 is not the answer. For the weather to be nice in this case, all 3 of your friend NEED to have lied to you. Therefor the odds are 1/27. It's really shocking to see how many people post incorrect answers here with such confidence. That said, Bayes' rule is somewhat counterintuitive if you're not familiar with probability theory. Let P(y|r) = prob of each yes given raining = 2/3, P(y|n) = prob yes given not raining = 1/3. Let P(r) = probability of rain = 1/4 given the prior knowledge. P(n) = probability of no rain = 3/4. P(r | y^3) = ( P(y^3 | r) P(r) ) / ( P(y^3 | r) P(r) + P(y^3 | n) P(n) ) = ( P(y | r)^3 P(r) ) / ( P(y | r)^3 P(r) + P(y | n)^3 P(n) ) = ( (2/3)^3 (1/4) / ( (2/3)^3 (1/4) + (1/3)^3 (3/4) ) = (2/27) / ( (2/27) + (.75/27) ) = 2/2.75 = 8/11 What if the answer is 50% since the chance of rain and not rain does not depend on what your friends tell you. In the absence of further information, the only correct answer is the posterior probability of rain p is in the interval (0, 1). In the absence of further information any prior is as good as any other, so by implication the posterior can take any value as well. The interval for p can be restricted to [0, 1] on the assumption that the question to the friends would not be posed if the prior is absolute certainty whether it will rain or not. With the further assumption that the prior probability is measured with limited precision (e.g. rounded to a percentage point), the posterior would be in the interval (0,075, 1). If the alternative assumption is made that information from the friends will be requested only if it had any chance to move the posterior below or above 0.5, the posterior interval for the probability is (0.5, 1). any more precise answer than that requires further information about the prior which is not supplied in the original problem formulation. Also note that even a precise answer about the probability of rain is not sufficient to answer the question whether an umbrella should be brought or not. Assume probability of raining in Seattle P(R) = 1/4 Assume friend says Y 50% of the time (Theoretical probability) P(Y) = 1/2 Probability of friend saying yes given its raining P(Y/R) = 2/3 Probability of 3 friends saying yes given its raining = P(YYY/R) = 8/27 Probability of 3 friends saying yes = P(YYY) = 1/8 P(R/YYY) * P(YYY) = P(YYY/R)*P(R) P(R/YYY) = 8/27*1/4/(1/8) = 16/27 (About 59%) A posterior probability of 59% given 3 yes and a prior probability of 25% sounds reasonable to me The probability of each of the friend say "YES" is 2/3 * 2/3 * 2/3 = 8/27. Now the probability that it is actually raining in Seattle depends on that how do I select them to phone. There is only three way to select and phone them. So, the probability that it is actually raining in Seattle is 3 * (8/27) = 8/9. Probability that it is raining given that all 3 of them said "yes" = P(AT LEAST one of them is telling the truth) = P(exactly 1 of them telling the truth) + P(2 of them telling the truth) + P(all 3 of them telling the truth) P(exactly 1 of them telling the truth) = P(of first person telling truth) * P(of 2nd person telling lie) * P(of 3rd person telling a lie) = (2/3) * (1/3) * (1/3) = 2/27 + P(exactly 2 of them telling the truth) = P(of first person telling truth) * P(of 2nd person telling the truth) * P(of 3rd person telling a lie) = (2/3) * (2/3) * (1/3) = 4/27 + P(exactly 3 of them telling the truth) = P(of first person telling truth) * P(of 2nd person telling the truth) * P(of 3rd person telling the truth) = (2/3) * (2/3) * (2/3) = 8/27 ANSWER: Probability that it is raining given that all 3 of them said "yes" = P(AT LEAST one of them is telling the truth) = P(exactly 1 of them telling the truth) + P(2 of them telling the truth) + P(all 3 of them telling the truth) = (2/27) + (4/27) + (8/27) = 14/27 Rule of conditional probability states P(A|B) = P( A & B ) / P(B) Reformulating to this case, P(Rain | 3Y) = P(R & 3Y) / P(3Y) P(R & 3Y) = 2/3 ^3 (if it is raining, then they must all speak the truth) = 8/27 (one could multiply probability of rain here. I assumed as prior) P(3y) = all truth or all lie = 2/3 ^ 3 + 1/3 ^3 = 9/27 hence P(R | 3Y) = 8/9 |

If you were hired, where would you want your office to be? 18 AnswersAt home. next door to the boss At home. Show More Responses next to you. Next to the boss - want the boss to see me and hear me. Want the boss to relie on me and see me as a great team player and someone that can be relied upon. I get an office? Sweet! Seriously, though, I'd want it to be near others who work on the same type of projects I do, so I can help and be helped accordingly. Next to the boss, but near enough to the team I work with to be well-integrated with them. Anywhere. It does not matter. It doesn't matter to me. Doesn't matter. My cubicle has always been Seat 10A on United Airlines... In a open cubicle, close to the team ! With a morale boosting view. Anywhere and everywhere... Show More Responses my car my own view is not mostly what you working but the impact you added to what you are doing I'd eventually like yours when I help you get promoted. Until then, I don't care where I sit as long as it puts me where I can make the most impact. I was not aware that Electronics Technicians had their own offices. I was under the impression that technicians were field employees, and thus, their office was their laptop in a van or at McDonalds. :-) In a large crowd of people so I could continue to receive feedback on behalf of work. Thus allowing me to better myself and the company. |

### QA Automation Engineer at BitTorrent was asked...

A dwarf-killing giant lines up 10 dwarfs from shortest to tallest. Each dwarf can see all the shortest dwarfs in front of him, but cannot see the dwarfs behind himself. The giant randomly puts a white or black hat on each dwarf. No dwarf can see their own hat. The giant tells all the dwarfs that he will ask each dwarf, starting with the tallest, for the color of his hat. If the dwarf answers incorrectly, the giant will kill the dwarf. Each dwarf can hear the previous answers, but cannot hear when a dwarf is killed. What strategy should be used to kill the fewest dwarfs, and what is the minimum number of dwarfs that can be saved with this strategy? 21 AnswersThere are 2 different strategies, each dependent on whether there are an even or odd number of white and black hats in play. The minimum number of dwarfs that can be saved with the correct strategy is 9. There is only one strategy, does not matter how many white or black hats they are. They are always 9 dwarfs that could be saved. You just have to know about XOR. Well... In my opinion all can be saved! The tallest dwarf can see 9 hats in front of him ( 4 white and 5 black hats or the other way around). Then he knows the color of his hat because there has to be 5black and 5 white hats. The next dwarf by the size just has to believe to the tallest dwarf that he is right-(and he is) In addition ,he can see 8 hats in front of him and when he adds the color of the tallest dwarf he knows which color he has. Again,the 3rd one in a row knows about 2 colors before and can see 7 colors in front of him.And when he adds 2 colors he already had heard of -he knows total number of 9 hats as well.So he just has to see which color is less presented and that is the color of his hat. And so on until the last dwarf...All saved Show More Responses the question does not mention that there will be equal number of white and black hats ! Since they do not know, how many black or white hats there are, the following strategy will save min 5 dwarfs: The first dwarf asked names the color of the hat of the 2. dwarf. He has a 50% chance that that's correct. Anyway, the second dwarf then knows, the color of his hat and names it correctly. the 3. dwarf again names the color of the hat of the 4th dwarf and has a 50% chance to survive, while the 4th dwarf can name the correct color of his hat. a.s.o. => all dwarfs with an even number will survive and all the others have a 50% chance Don't over-think it. Each dwarf simply has to state the color of the hat worn by the dwarf directly in front of him. The tallest would have to sacrifice himself in order to save the other 9. Easy. The tallest is the only one that cannot be saved so instead of trying to guess his color, he yells out a sequence of letters starting from the shortest like BWBBBWWBB. Next :) Your going to definitly get 9 because when the dwarves collude the tallest tells the next tallest his hat color and so on down the line. Now you have a 50/50 chance that the tallest will get his color correct and live so you have 9 with a 50% chance of 10 Each dwarf can state the color of the hat worn by the dwarf in front of him, but the thing is, that color may not be the color of his own hat.So he may be killed by the giant.In that case, as mentioned above all odds should tell the color of next so that all even number will be alive and they have 50% chance of surviving. Think broadband communication. Exploit the capabilities of the communications medium. A minimum of nine dwarves can be saved based on the information provided in the original post I viewed. The strategy is for each dwarf to employ the expected language to communicate the color of their own hat to the giant, while simultaneously employing a vocal pitch protocol to indicate the color of the hat of the dwarf in front of him, high pitch for white and low pitch for black. The original post, indicates the dwarves may collude prior to the distribution of hats, so there is opportunity to negotiate such a simple broadband communication protocol. The tallest dwarf only has a 50/50 chance since the number of black and white hats in play is not known (rhetorical question, what are the odds the tallest dwarf's hat is black if he turns to find that all nine hats in front of him are white? I don't know, but odds are high that the giant is a sadistic bloke). The original post I viewed is here. http://www.businessinsider.com/toughest-job-interview-questions-2013-7#a-dwarf-killing-giant-lines-up-10-dwarfs-from-shortest-to-tallest-each-dwarf-can-see-all-the-shortest-dwarfs-in-front-of-him-but-cannot-see-the-dwarfs-behind-himself-the-giant-randomly-puts-a-white-or-black-hat-on-each-dwarf-no-dwarf-can-see-their-own-hat-the-giant-tells-all-the-dwarfs-that-he-will-ask-each-dwarf-starting-with-the-tallest-for-the-color-of-his-hat-if-the-dwarf-answers-incorrectly-the-giant-will-kill-the-dwarf-each-dwarf-can-hear-the-previous-answers-but-cannot-hear-when-a-dwarf-is-killed-the-dwarves-are-given-an-opportunity-to-collude-before-the-hats-are-distributed-what-strategy-should-be-used-to-kill-the-fewest-dwarfs-and-what-is-the-minimum-number-of-dwarfs-that-can-be-saved-with-this-strategy-11 So reading the answers provided they all have some assumptions e.g. that there are as much white hats as there are black hats. I think that Christian's comment on aug 13-2013 was very close but I'm thinking about communication integrity confirmation techniques. One of them is using a parity bit to confirm the message was correct. This could be applied right here to save at least 9 with a 50/50 chance of saving the 10th (and tallest dwarf). I will explain it but for ease of explanation I will use binary 1 and 0 for black and white. Number 1 being black hat and number 0 being white hat. Let's say we got a (random) hat sequence of 0001011101 with the tallest dwarf on the right and the shortest on the left. While colluding prior to the distribution of hats the dwarfs agree upon even or uneven parity. This means the total amount of 1's (black hats) must be even or uneven including the parity bit. In this case the 10th dwarf will count as parity bit. So if we'll take an even parity, the number of 1's must be an even number in total. When the questioning starts the tallest dwarf will see the hats in front of him being 000101110. The tallest dwarf counts four 1's (black hats) so to make the parity even he has to say 0 (white hat). He will get killed but the dwarfs in front of him will know the parity bit is 0 so the 2nd tallest dwarf will see the hats in front of him as 00010111. He will also count 4 and knowing that the dwarf behind him said 0 he'll know that the total amount of 1's is an even number thus concluding he has a 0 (white hat) and will state he has a white hat. Same goes for the rest of the dwarfs and so 9 will be saved. The 10th would've been saved it the dwarfs agreed on an uneven parity. That's why there's a 50/50 chance the 10th will be saved. I'm pretty confident this is the answer but if you want to understand it better (maybe my explanation is a bit vague) go search for "parity bit" on the internet. @Kristen - you have so underthought it! What if the dwarf behind you says "black" and the dwarf in front of you has a white hat???? @JustJanek- Your solution is close, but not correct. Every dwarf needs to consider the parity of a number composed of all the following dwarfs and all the dwarfs behind. The first dwarf says the parity of the 9-bits number in front of him. Assuming that all the other dwarfs know the trick and they stay alive, each dwarf needs to compare the initial parity with the parity of an 8-bit number composed of the hats in front of him and the hats behind (assuming that the dwarfs behind gave the right answer), if the parity is the same, he knows that he has a white hat, otherwise he has a black hat. Show More Responses You can save 9 dwarves at least. Dwarves agree that the tallest one says he is wearing black hat if he sees odd number of black hats in front of him and he says white hat if he sees even number of black hats in front of him. So, the tallest one has 50-50 chance of survival and other dwarves survive 100%. What is the minimum number of dwarfs that can be saved with this strategy? 9 First of all, let's numerate the dwarfs as N1, N2, N3, etc. with N10 being the tallest. Now, N10 will state the color of N9 as his own answer, "My hat is WHITE". Based on this answer, N9 will state his color with a positive statement if the color of N8 is the same as his, "My hat is WHITE". Based on N8's answer, N8 knows that his color is WHITE, now, he will state his color depending on N7. Let's say N7 is black, so N8 will state, "My hat is NOT BLACK". N7 knows that his color is BLACK, but N6 is white, so he will use a negative statement, "My hat is NOT WHITE" and so on. Full example: N10 = BLACK N9 = WHITE N8 = WHITE N7 = BLACK N6 = WHITE N5 = WHITE N4 = WHITE N3 = BLACK N2 = BLACK N1 = WHITE N10: My hat is WHITE (Dies) N9 = My hat is WHITE N8 = My hat is NOT BLACK N7 = My hat is NOT WHITE N6 = My hat is WHITE N5 = My hat is WHITE N4 = My hat is NOT BLACK N3 = My hat is BLACK N2 = My hat is NOT WHITE N1 = My hat is WHITE N10 will have a 50/50 chances of survival... I'm sorry N10, I couldn't save you :'( What is the minimum number of dwarfs that can be saved with my strategy? - 10 My strategy doesn't make any unmentioned assumptions (such as equal number of white and black hats, etc). At the same time, it doesn't add any unmentioned constraints either. The strategy is simple to the point of appearing simplistic. But it meets all the requires. The strategy is: When asked, every dwarf answers "Not RED". This is not an incorrect answer and the Giant, if he were an honourable giant that is :), would have to let the Dwarf live. On a different page altogether, I went through all of the above answers. Not to sound patronizing, but some of the solutions were quite brilliant. I was thinking if I could even come up with that even after years of pondering. But I must admit, all of the above strategies are made by an outsider (ie. us) who is not impacted by this fate. Whereas in the casestudy, the strategy has to be devised by the 10 dwarfs, who face the impact of this strategy. Not to bring in factors such as emotion, the bell curve distribution of intelligence, and other such anal considerations. But I thought it was important to bring in Game Theory, and that all Dwarfs are rational, and that all rational people do not want to harm themselves in any way. In other words, when a strategy's success depends on the conformance to that strategy by ALL the participants, the strategy should also benefit ALL the participants if it is to succeed (or in the least, should not harm even ONE participant). In all the above strategies, since even in the best case scenario the poor Tallest Dwarf has only a 50% chance of living, can we assume that he would conform to this strategy? Each dwarf should pronounce color of hat of dwarf before him. This way they can save atleast 9 dwarf out 10. Well, Once the dwarfs are lined up in descending order. Without any kind of assumption, 9 people can be saved with a 50% probability of the 10th (tallest). Here is how it can be achieved, The strategy is to call the color of odd number of hat. Say for example, the tallest dwarf sees 3 Black and 6 whites, it will call out Black(it may or may not die with 50% chances). Now, the 9th tallest dwarf knows what the 10th dwarf could see and if it (9th) dwarf sees the same odd number of black hats, it will know it has white hat. Next, 8th dwarf knows number of odd (black) caps 9th dwarf could see and if it(8th) finds 1 less black cap, it would know it is wearing a black cap.. and so on. 10- White (calls out Black because it sees odd number of black hats) -dies(assume) 9- White (calls out White because it could still see 3 black hars) 8- Black (sees that there is one less black hat as mentioned by 9th, hence identifies that is wearing black) 7-White (calculates that 8th is wearing black and he could see 2 black, hence identifies itself wearing whiteO( 6-White 5-White 4- Black 3-Black 2-White 1-White Sorry if there is confusion in the way I have answered. all can b saved... They will exchange their hats in Circular form...person10 can see the person1 color of hat and after exchange every dwarf will tell color to their previous ones and person 10 knows the color before changing it from dwarf 1. d10->d9->d8->d7->d6->d5->d4->d3->d2->d1->d10 CIRCULAR EXCHANGE OF HATS First lets look at number of back and white hats...To satisfy the condition "black and white" hata there is minimum one black or white hat present. So its minimum (9 black & 1 white) or (1 white & 9 black) hat being distributed randomly amoung dwarfs. The story says dwarfs are alowed to speak before execution. Lets make a strategy of saying only one colour before execution eg) black or white. Minimum probability of (9 black & 1 white) hats and all the dwarfs say "white"...In this case one is saved but all nine dead. If all dwarfs say "black"..Nine are saved but one is dead. The same applies for the minimum probability of (1 black and 9 white)hats. Thus minimum one dwarf is saved and maximum nine dwarfs can be saved. If each dwarf say the colour of hat in front of him..(dwarf can hear previous answer) then at least five people are saved. |

### Trader at Jane Street was asked...

What is the sum of the digits of all the numbers from 1 to 1000000? This is different from the sum of the numbers. For instance the sum of the numbers from 1 to 10 is 55 whereas the sum of the digits is 46. 16 AnswersThe main idea is that if you write all the numbers from 0 to 999999 down as six digit numbers (possibly prepending zeros) then all digits appear the same number of times. So, its digit appears exactly 6 x 1000000/10 = 600000 times. so the result is 600000x 45 +1 (+1 for the number 1000000) Maybe I'm reading it wrong, but isn't the sum of the digits of 1-10 just 11? 1 digit for 1-9, 2 digits for 10-99? So it's 9+90*2+900*3+9000*4+90000*5+900000+6 no, I think the easy way to solve it in your head is to remember that when adding all digits 1 to 100, you have 50 pairs: 1+100, 2+99, etc. Each pair is 101, times 50 is 5050. 1 to 1000 would be 500500 so 1 to 1,000,000 would be 500,000,500,000 Pretty cool, huh? Show More Responses scott, dude you should add digits not the numbers, so 99+2 = 18+2 =20. not 101 Scott's answer from May 24 is the correct way to think about it if summing the numbers. The answer is 27,000,001 - if you do it programatically the operation is a simple map reduce - simply map a digit sum function across the list of values [1, 1000000] and then reduce an addition operator across the result. Python Proof: >>> sum(map(lambda n: sum(map(int, str(n))), xrange(1, 1000001))) 27000001 21085156 Don't mind the above answer. I read the question wrong. Each digit will appear 1+10+100+1000+10000+100000 times. so the answer is 111,111*45+1=4444440+555555+1=4999996 nb is right. Another way you can think about it easily is if you want the digits from 1 to 1,000,000 then each digit should appear 1/10 * 1,000,000 times. So 100,000 x 6 (for each place value) x 45 (the sum from 1 to 9) + 1 (for the 1 million) 27,000,001 I think John Doe is right. 27,000,001 is what I got. Think of each number as a 6 digit number. The average number each digit could be from 000,000 to 999,999 is (9+0)/2=4.5. Since the average of each number is 4.5 and there are 6 digits the average sum of the digits for a 6 digit number should be 4.5*6=27. There are 1 million numbers from 000,000 to 999,999 so the sum of the digits from 000,000 to 999,999 is 27,000,000. Subtract the digits of 000,000 which is just 0 and add the digits of 1,000,000 which is just 1 to get 27,000,001. For arbitrary a=1,...,9 accounts the how many a's appear in the six digit number. 1 * C(6,1)*9^5 + 2* C(6,2)*9^4 + 3 * C(6,3) * 9^3 + 4*C(6,4)*9^2 + 5*C(6,5)*9 +6*C(6,6) = 600000. 600000*(1+2+3+...+9)= 27000000. Then add the 1 from 1000000. The ans is 27000000. Show More Responses 2.7 x 10^7 + 1 general rule: to 10^n answer is 10^n * n * 4.5+1 here n=6, result is 27 000 001 1-1000000 same as sum of digits in 0 - 999999 then plus 1, treat 0 - 999999 as a 6 digit random number, then the digit of sum is sum of digit for each number: 1000000*(4.5*6)=27000000, so answer is 27000001. |

### Software Engineer at Google was asked...

One independent survey showed that 70% on people asked like coffee. Another independent survey showed that 80% of people like tea. What is the upper and lower bound of peoples who likes both coffee and tea 14 AnswersUpper: 70% Lower: 50% 20, 80 50, 70 Show More Responses It is 70% Both High & Low. Any higher and people don't like Coffee, any lower and they don't like coffee. This is assuming 0% margin of error and a perfect survey. Take 100 people. 1-100. Assume, 1st 70 likes coffee (1-70) and 1st 80 likes tea(1-80). So, atmost 70 can like both tea and coffee. Now, take the same 100 people. Assumed, 1st 70 likes coffee and last 80 likes tea(21-100). So, that make atleast 50 people likes both coffee and tea. So, upper is 70 and lower is 50. Only 70% of people like coffee, so that would be the upper bound since they need to like both coffee and tea. 80% of people like tea, which leaves 20% that don't like tea. Once again, they need to like both coffee and tea, so you subtract the 20% that don't like tea from the upper bound which leaves 50% for the lower bound. This assumes the survey was 100% accurate and had 0% margin of error, which means they would have had to interview the same 100 people in both surveys and the results are only valid for the 100 people surveyed. Only 70% of people like coffee, so that would be the upper bound since they need to like both coffee and tea. 80% of people like tea, which leaves 20% that don't like tea. Once again, they need to like both coffee and tea, so you subtract the 20% that don't like tea from the upper bound which leaves 50% for the lower bound. This assumes the survey was 100% accurate and had 0% margin of error, which means they would have had to interview the same 100 people in both surveys and the results are only valid for the 100 people surveyed. The fact that these are two independent surveys suggests that they were not conducted on the same group of people. Therefore, if we were to extrapolate this survey data for people in general, you can only conclude that the upper limit is 70% but the lower limit can be zero since 100% of the set of people who like coffee may not like tea. Seventy percent of people like coffee so the upper bound is limited by that seventy. 80 percent of the people who like tea multiplied by 70 percent who may like coffee is 56 percent, so 56 lower 70 upper. This problem reminds me of a convolution operation. Convolve 1111111000 with 1111111100 and you sortof get your answer. Assuming the two independent surveys have been done on the same sample of people, following a consistent set of rules: n(AUB) = n(A)+ n(B) - n(A intersection B) Assuming a population of 10 people: 10 = 7 + 8 - x or, x= 5 or, 50% = lower limit And of course the max will the min(coffee lovers, tea lovers) = min(7,8)= 7 ~ 70% Rock has it right. The trick is that it is; two independent studies, not we asked the same group. 70% is the easy part but of those 70% that like coffee 80% like tea so that means at least 56% like both Given that they were the same set of people. Or assuming that the sample size was large and a representative of entire population, the lower bound is 50% and the upper bound is 70%. Here's how: Let the population be represented by 10 dashes: Worse Case: - - - - - - - - - - (total) - - - - - - - (70% coffee) - - - - - - - - (80 % tea) If you take the common of above two, you get 5 dashes. Meaning 50% atleast like both coffee and tea Best case: - - - - - - - - - - (total) - - - - - - - (70% coffee) - - - - - - - - (80 % tea) Which gives you 70% Show More Responses Key is "two independent surveys". Also note that it is not mentioned anywhere that both surveys had options of coffee and tea or they were the only options. So, lets say there are 200 people of which 70 like coffee and 80 like tea. We don't anything about remaining 50. They could like liquor for all we know. Maximum liking both tea and coffee: 150 out of 200 i.e. in survey one who said they like coffee also like tea and similarly in survey two who said they like tea also like coffee. Hence, 3/4 or 75% is the upper limit. Lower bound: 0%. None of them liking coffee like tea and vice-versa. Remaining ones like coconut water :) |

### Software Engineer In Test at Google was asked...

How would you determine if someone has won a game of tic-tac-toe on a board of any size? 14 AnswersCheck all rows, check all columns, check two diagonals. If there exists a diagonal, row or column of 'X' or 'O' someone has won the game. For a board of size N^2 runtime is N. There's a way to do this in constant time... I think maybe this question is worded a bit wrong, because given a tic-tac-toe board you would need to read in at least some of the values on the board to figure out if someone has won, and this would be impossible to do in constant time (the larger the board, the more values you would have to read). I think they must mean how can you determine if someone has won during a game in real time, as in checking after every move. This can be solved with a strategy in constant time. My solution would be: Create an array of size 2n+2 at the beginning of the game and fill it with zeros. Each spot in the array will be a sum of X's or O's horizontally (the first n places in the array), vertically (the second n places in the array) and diagonally (the last 2 places). Then with every move, you add 1 to the 2 places (or 3 if on a diagnol) of the array if X, and subtract 1 if its an O. After adding you check and see if the value of the array is equal to n or -n, if it is, n mean X has won and -n means O has won. I would bet there is a more elegant solution than creating a large array, but since this isn't my job interview I can't be bothered trying to figure one out. :) Show More Responses You would determine the winner by identifying the first player to string together three consecutive X's or O's. The size of the board is irrelevant. You are looking for the winner. The easy and obvious answer is to check the happier player, not the board. While this may ignore the "engineering" side of the question, it does demonstrate the logic of searching for the simplest answer becuase that solution will be the same regardless of the board configuration. Happier player doesn't always mean the winner. A father teaching his son how to play tic tac toe for instance could be happier if his son actually beat him at the game. Your "simplest answer" is wrong. Assume that you are handed a board with no prior knowledge of what has happened in the game. Assume that, to win on a board of size NxN, the player must have N 'X' characters or 'O' characters in the same row, column, or diagonal. Assume that, for our problem, we are only checking if the winner is 'X'. We have to make at least one pass through the game board, but we should be able to solve the problem in one pass without checking any cell twice. Target running time O(N^2) for a board of size NxN. boolean checkXWinner(int[][] a, int n){ int[] diagonalSums = new int[2]; int[] columnSums = new int[n]; initialize diagonalSums and columnSums with zeroes; int rowSum = 0; for (int i = 0; i < n; i++) for (int j = 0; j < n; j++) if (a[i][j] = 'X') rowSum++; columnSums[j]++; if (i == n-1 && columnSums[j] == n) return true else if (i == j) diagonalSums[0]++; if (i == n-1 && diagonalSums[0] == n) return true else if (i = n-1-j) diagonalSums[1]++; if (j == 0 && diagonalSums[i] == n) return true if (rowSum == n) return true (sorry - forgot last line, put this after if (rowSum == n)....) else rowSum = 0 // Need to reset the row counter Ask. Count the number of times X and O appear on the board. Whichever has the greater count is the winner. there can be more than one winner @count : The game can be tied even though one has greater count. iterate the board and every time you find a players symbol peek forward if the board contains other two symbols at correct indices (there are four feasible patterns). this is in constant time. Show More Responses I came across this question on a Google search for something else related to tic-tac-toe. I was just moments ago thinking of this exact problem (fastest way to determine if game is in "won" state) and I'm surprised I have not seen it here... Why check every square??? Assuming an NxN square board, ANY winning arrangement MUST include a square on the diagonal. Iterate over the diagonals, and recurse both vertically and horizontally for matches, breaking when a non-match is found. On the center square check both diagonals in addition to the vertical and horizontal. The only scenario that requires all squares to be checked is if one edge is a winning edge and even then there's a few constraints required to force it. |

The Game of Nim worded diffently. 11 AnswersThe Name of the Game. The anwer is 'Take' from the german word nimm. There is a game called 'The Game of Nim' that has a specific mathematical equation that must be utilized in order to win the game. Nimm is the German word for Take, so you must figure out the best way to take the matches without your opponent beating you at it. Show More Responses The Game of Nim is a simple board game in which you and your opponent take turns removing a number of matches from one of the rows (normally about 5 rows) of matches on the board. The person to take the last match off the board is the winner. The reason why it is of interest to us as prospective software engineers (and why you probably asked this question) is that it has some interesting binary number properties making it fairly trivial to write computer code to ensure a win every time (every time there is a starting advantage, that is). Would you like me to go into more detail? Ok, well in brief then, basically the trick is to take the number of matches in each row and represent this as a binary number. Then, either by hand or with a program, do an Exclusive Or operation on the numbers. Then whenever you take some matches, just ensure that the remaining total is always zero after your turn and you will be sure to win by the end of the game. Maybe I should also add (and I'm thinking out the box here), that sometimes we as people are up against a challenge or opponent where succeeding or beating them is seemingly reliant on chance or luck. However, with careful analysis of the problem and good strategising, it turns out it is actually possible to ensure success just about every time. On the other hand, there are times when the odds are against us from the start. Then either we must stand up for what we believe is fair (i.e. be aware and vocalise that we cannot possibly win), or else acknowledge that our opponent is worthy and will ultimately get the better of us. Yet it should be noted that we can still stay strong and be competitive from the beginning allowing us to possibly take advantage of any mistakes or weaknesses our opponents or challenges might display. That is the Game of Nim worded differently. What does this question have to do with Quality? Nim's Game If I was interviewing you and asked you that question, I would be trying to determine if you could take a simple problem and provide a simple solution. If you went off into the weeds like Andrew_Bryce did, I would be wondering how effective you would be solving tons of simple issues. Also, if you answered the wrong question (what is the Game of Nim?) and not the question I asked (how would you word differently the phrase The Game of Nim?), I would be wondering how good your communication skills were. If I were being interviewed, I would wonder about the interviewer's communication skills. "The game of Nim worded differently" isn't a question. It isn't even a sentence. "Foaming Theme" 1) I cheated 2) I didn't even know what Nim was before I looked it up Here's how I would interpret some answers and the job I would recommend for them Anonymous: Huh? I think you're trying to be a smartass, but I don't get it - Cafeteria Worker Ryan: Knowledgeable - Content provider SelenityHyperion: Knowledgeable, informative and relatively succinct - Writer Andrew_Bryce: Detail oriented and a perfectionist - Software tester, some forms of coder Astrochimp: Focused - Project Manager OneEye: Thinks his answer is the only correct one - Clearly VP material Count Negroni: Nit Picker - Editor I don't think there's supposed to be one correct answer Ha ha, jokes on me. "The Game of Nim worded differently." is not the actual question, just a vague description. I would answer: "Are you talking to me?" Because it sure sounds like you're high. Then I'd get up and leave. Strategic domination |