# Brain teaser Interview Questions

interview questions shared by candidates

## Brain teaser Interview Questions

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

Write a function Brackets(int n) that prints all combinations of well-formed brackets. For Brackets(3) the output would be ((())) (()()) (())() ()(()) ()()() 11 Answerspublic class Parenth2 { static int total = 3; static private void Brackets(String output, int open, int close, int pairs) { if ((open == pairs) && (close == pairs) && output.length() == total * 2) { System.out.println(output); } else { if (open < pairs) Brackets(output + "(", open + 1, close, pairs); if (close < open) Brackets(output + ")", open, close + 1, pairs); } } public static void main(String[] args) { Brackets("", 0, 0, total); } } You almost got thte answer, just a couple of errors... /* * To change this template, choose Tools | Templates * and open the template in the editor. */ /** * * @author Owner */ public class Parenth4 { static private void Brackets(String output, int open, int close, int pairs, boolean opened, boolean closed, int total) { if ((open == pairs) && (close == pairs) && output.length() == total * 2) { System.out.println(output); } else { if ((open < pairs)) Brackets(output + "(", open + 1, close, pairs,true, closed, total); if ((close < open)&& opened) Brackets(output + ")", open, close + 1, pairs,opened,closed, total); } } public static void Brackets(int total) { Brackets("", 0, 0, total,false,false, total); } public static void main(int number){ Brackets(number); } } This is a DP/memoization question, I believe. The base case is 0 and 1 bracket (the answer is empty or (). The recurrence is: bracket(n) = all combination of results from bracket(n-1) and from bracket (1) from bracket(n-2) and bracket(2) . . from bracket(1) and bracket(n-1) lastly, '(' . bracket(n-1) ')' The DP version of the above recurrence is straightforward. (Btw, this recurrence obviously will produce duplicate, but it's not hard to produce a modification that does not produce duplicate.) ;) Show More Responses Here is a F# implementation: let rec Br total output openp closep = if openp = total && closep = total then printfn "%s" output else if openp < total then Br total (output + "( ") (openp + 1) closep if closep < openp then Br total (output + " )") openp (closep + 1) let Brackets total = Br total "" 0 0 Brackets 5 let read = System.Console.ReadLine() void parens(int nPairs) { parens("", nPairs, nPairs); } void parens(string ans, int leftCount, int rightCount) { if (leftCount==0 && rightCount==0) { cout 0) { parens(ans+"(", leftCount-1, rightCount); } if (rightCount>leftCount) { parens(ans+")", leftCount, rightCount-1); } } To Remove duplicates simply use java Set :) public static String bracket(int a) { Set s = new TreeSet(); bracketIntern(s, a, "", ""); System.out.println(s); return s.toString(); } public static void bracketIntern(Set set,int a, String preFix,String suffix) { if(a == 1) { set.add(preFix+"()"+suffix); return; } bracketIntern(set, a-1, preFix+"()", suffix); bracketIntern(set, a-1, preFix+"(", ")"+suffix); bracketIntern(set, a-1, preFix, "()"+suffix); } I think you can also build a trie, and the traverse the trie to print out all the combinations. Complete python program to print the all combinations of well-formed brackets. How to calculate its Big-o? The recursive recurrence seems a little bit complicated. ---- def brackets(n): sol = [] li = [' ' for x in range(n*2)] def recu_brackets(opened, closed): if n - opened: li[opened + closed] = '(' recu_brackets(opened + 1, closed) if n - closed and opened > closed: li[opened + closed] = ')' recu_brackets(opened, closed + 1) if opened == n and closed == n: sol.append(''.join(li)) recu_brackets(0, 0) print ' '.join(sol) brackets(3) wasn't all that neat. o well public class MakeBrackets{ static List make(int number){ List list = new LinkedList(); if (number == 0) {list.add(""); return list;} if (number == 1) {list.add("()"); return list;} for (int i = 0; i < number; i++){ for (String item : make(i)){ for (String item2 : make(number - 1 - i)){ list.add("(" + item + ")" + item2); } } } return list; } public static void main(String[] args){ System.out.println(make(Integer.parseInt(args[0]))); } } def brackets(n): return bracketsPrefix("", n, n) def bracketsPrefix(prefix, opens, closes): total = 0 assert opensopens: total += bracketsPrefix(prefix+")", opens, closes-1) if opens>0: total += bracketsPrefix(prefix+"(", opens-1, closes) return total if __name__=="__main__": for i in range(6): n = brackets(i) print i,n x = raw_input("** ") This question gets so popular and I found this article has a really good explanation: https://goo.gl/E0m0EC |

### Software Engineer at Linden Lab was asked...

An abstract question about ways to simulate behavior of workers who move stools around a workspace in some controlled way, and what the expected results would be. 12 Answerskkkk Answer__ Who wants to have a race through the office? Anyone who wants to participate stand on this side of the room, anyone who wants to bet on the races outcome stand on the other. Brownian motion. Ant hill. Dense flock of birds or school of fish. Show More Responses I would simulate it with dance--either like "The Nutcracker", very graceful-like, with a very harmonious outcome; or like Michael Jackson's "Thriller", very zombie-like, with excitement as your outcome! Music added to the environment could result in teamwork developing and improved results from working together to a common known beat. An ideal workspace does not require that much movement where is the question? I don't see a question to answer. I suggest that the question be rephased. We are to make up the question. That's what we are being asked to do. What is the way to confuse a potential new-hire with incomplete chairs in a confined space? And I'd like to wager all of it! this question was the most ridiculous out of the entire gauntlet. The answer was dependent on knowing the layout of the old office space that they haven't been in for 5 years now, and the office culture in solving issues through desk-visiting rather than meetings in meeting rooms, because the CEO (2 CEOs ago) severely frowned upon excessively-staffed, expensive meetings. The answer is "the stools end up on the edges of the room". |

### Associate at Capco was asked...

Let's say you have 10 bags of nails, and each nail weights 10g. You also have a scale upon which you can weigh a combination of the bags or nails from the bags once. You know that one of the bags is off by a certain amount of grams, but you don't know which bag or by how much. How do you find out, with one weighing, which bag it is? 10 AnswersNumber the bags 1 through 10. Take 1 nail from bag 1, 2 nails from bag 2, and so on. Take those 55 nails, which should weight 550g, and place them on the scale. If the weight differs from 550g by a certain amount, then that tells you which bag it is and how much it's off by. If each "nail weights 10g" as posed in the problem the ONLY weight you will get is 550g. My Idea is the instructions state "can weigh a combination of the bags or nails from the bags once." Place bag on scale result should = 100g, add bag 2 = 200g ect. when you add a bag and the result = less than the 100g, result = bag number and weight and difference. As you have continually adding weight is counts 1 time. Who cares? If you run out of nails go get some more. Show More Responses You guys are doing this wrong. This is a twist on a very old problem, that actually has to do with twelve balls and can use the scale only three times. The trick is to divide the bags into smaller groups of 3, and see in which direction the bags move. It's easier if you draw it out. It is from "games for the super intelligent" by James Fixx. If I can only weight them once... then I'd put all the bags on, one weigh in and take the bags off one by one. The weight should go down by 10g each time until you get to the bag which is off. Technically it's one weigh in at the beginning and the rest are clearing the scale. Take all the nails out of the bags weigh them divide by 10. The number wont come out even, bag up your nine even piles, then the last bag you have is the one that is "short". This is a stupid question to use in an interview. Presumably by the time people get to this level, they've already taken the SAT and GRE, the latter of which includes a logic section. Test people on things that are going to matter for the company's bottom line, not a logic problem. I agree with kgar 79. The problem states that you can weigh the nails individually, which means you can take them out of the bag. Once you do, you'll be able to see which bag has more or less nails than the others. Since the nails all weigh the same, any variation in weight, will be determined by whole nails. Simple, and it can be done without weighing anything. Count the number of nails in a bag and multiply the number by 10g. kgar79 got it right. Just dump out all the nails, put the same number of nails in nine of the bags,their weight is now 10g x the number of nails per bag, then put the rest of the nails in bag #10. Then weigh it and now you know which and by how much it is off. You don't actually even need the scale. |

### Sales Strat Intern at Goldman Sachs was asked...

Suppose we hire you, and you and the rest of the new interns decide to go buy a cup of coffee. Each intern purchases one cup of coffee. One of the interns suggests everyone play a game. Everyone will flip a fair coin, dividing the group of interns into two subgroups: those that got heads and those that got tails. The game is this: whichever group is smaller evenly splits the cost of everyone's cup of coffee (i.e. if there are 5 interns, 3 get H, 2 get T, then the two interns that got tails each buy 2.5 cups of coffee). However, nothing says you need to play this game. You can choose to buy your own cup of coffee and not play the game at all. The question: Should you play this game? (Note: You may assume that there is an odd number of interns, so there are no ties, and that if everyone gets H or everyone gets T, then everyone loses and just buys their own cup of coffee). 18 AnswersHint: Despite its look, this is not a math question. I dont get it, Would you please provide more hints? Assume each coffee costs $1, for simplicity. So this is effectively a choice between two outcomes: paying $1 with probability 100%, or paying $0 with some probability and paying more than $0 with some probability. So you ask yourself: what is your expected cost in the second case? Give that a try and see if you can figure it out. However, I want to remind you that the question is "should you play this game?" The answer to this question isn't just a math question. If you only work out expected values, you've missed the point. For example, a separate question (with the same kind of flavor as the direction I'm trying to lead you) is this: suppose I give you a choice of two outcomes. Either you get $1 with 100% probability or I give you $500,000,000 with probability 1/100,000,000 and 0 otherwise. Which would you pick? Now what if it was the same first choice, but the second choice was $50 with prob 1/10 and 0 otherwise? Now which would you pick? These are the kinds of things you want to think about while answering this kind of question. Let me know if you have anymore questions. And if you want me to post the answer, just let me know. Show More Responses I think I get it. It's about investor's risk appetite. Investors is likely to take guaranteed gains, here is $1. Well yes and no. This is indeed a risk aversion question. If you work it out, you'll find that the EV in each case is exactly the same (your EV is -1 cup of coffee in both scenarios) but that's not the end of it. It's also really a question for them to test your risk aversion. You can really support either answer, and *should* comment on the validity of either answer. My answer was to go with buying my own cup of coffee, and followed it up with a story where a friend of mine had tried to get us to play credit card roulette (which is similar in spirit to this game) and I told him that I did, in fact, say no in that instance and why I said no. However, traditionally people are risk-averse when faced with gains and risk-seeking when faced with losses, so many would probably choose to play the game. But this is as much a question about your psychology as it is about your math skills. And that's why this is such an awesome question, and is probably a question that kills most people they ask it to. Would you please explain to me how do you get -1 for the second scenario? There are 50% H and 50% T. Therefore players have 50% opportunity in the winning group. Given 5 interns, there are two combination of lossing group (4 vs. 1 and 3 vs. 2). EV=0.5*0 + 0.5*(0.5*-5 + 0.5*-2.5) = -1.875. The probability of winning isn't 50%. It's actually slightly above 50%, but that's not the way to look at it. The total number of coffees that need to be bought is n, where n is the number of interns. Going into this game, every intern is the same so they each have the same expected value, and the sum of the expected values must equal -n. So everyone has an EV of -1 as claimed. Thank you very much. I finally got it. I'm still not sure why you said no if the EV is the same? It's a matter of risk preference. See the example I gave in my Mar 19 posting: "Suppose I give you a choice of two outcomes. Either you get $1 with 100% probability or I give you $500,000,000 with probability 1/100,000,000 and 0 otherwise. Which would you pick? Now what if it was the same first choice, but the second choice was $50 with prob 1/10 and 0 otherwise? Now which would you pick? These are the kinds of things you want to think about while answering this kind of question." You would probably not play in the first instance and consider possibly playing in the second. (And those instances even have the risky game with a HIGHER EV). This coffee game is the same kind of game, and even if the EV is the same in each case, the volatility is not the same. It is usually a good rule of thumb to take the lower volatility outcome if the EV is the same (think Sharpe Ratio here! Or think efficient frontier here! Both get the point across I think.). Does that help? I say yes. To play the game. Only because I'm prepared if I lost. The fact that I can afford to loose, makes me want to try my chance at wining I would play the game. Consider the expected price with n total interns splitting a total cost of P. That is P/2*E[1/(N)|you're paying]. This is simply equal to P/2*E[1/(N+1)] where n is a binomial now corresponding to n-1 interns. Notice that 1/n+1 is a concave function. That means that P/2*E[1/(N+1)] <= P/2*(1/E[N]+1)=P/2 * 1/( (n-1)/2 + 1)) = P / (n +1). So it pays to play on average Apologies, but both of the above answers are incorrect. The EV of playing is the same as the EV of not playing. Show More Responses No way would I play! The most money I save is a dollar the most money I can lose ( in the case of five interns) is 4 dollars... That's a 400 percent downside verse a 100 percent upside ( kind of you cannot technically compute your return on zero dollars invested). Plus, I do not even drink coffee! Stupid...I don't have time to play games at work. The EV in both scenarios are not the same: it's clear when think about the case where there are n = 3 interns. I've already explains why they are the same. In either scenario there will be n cups of coffee bought (3 in your case) so total EV is -n (-3 in your case). In the game each person is the same as any other so their EV must all be the same. That is why in the game the EV is -1 (same as if they buy their own cup of coffee). To argue otherwise is to argue that either the total EV is not the number of coffees purchased or that someone has an unfair advantage in the game. Incidentally, if you are going to claim that someone who has given you the answer is wrong, you should provide more of a response than "go think about it." if v = pay by game -1 . Then all have five intern have same distributed V by symmetry 5EV=0 It's zero sum. |

### Trader at Morgan Stanley was asked...

What is the square root of .01? 8 Answers.3167 - Just memorize it, it's very common. I think you mean the sqrt of 0.1, not 0.01. The sqrt of 0.01 is 0.1. Not to be a fuddy-duddy, but .3162 would be closer. (If you're going to take it to 4 decimals then you should use the closest 4th decimal.) Show More Responses Answer: 0.1 assuming my 2 SCI calculators and my manual calculations are correct Comments: 1- 0.3167 is the square-root of 0.1 not 0.01 2- Don't tell you interviewer he is wrong. You want the job don't you? 3- You actually need to use a 5/4 round-off using the 5th place for a more accurate 4th ( I think that's what you meant to say) 0.01 is the same as 1/100 and if you take the square root of the top and bottom, you get 1/10 or 0.1. The square root of 0.01 is neither 0.1 nor 0.3162. The square root of 0.01 is +0.1 AND -0.1 (Negative 0.1). The negative root is frequently overlooked. Isn't it prime already? So its sq rt is 0.01. I'm so confused! actually, square roots are always positive. the definition of a square root is that it's an absolute value. i.e. sqrt of (-1)*(-1) = |-1| = 1 My real response to this topic though is that I dread such silliness in interviews. In my current position I was asked a stupid question in a 5-on-1 interview and basically said what I thought about the question but still got the job - so maybe that was the correct answer. Hope I don't ever find myself desperately needing a job and having to endure this silliness again. |

### Systems Engineer at Google was asked...

How many trailing zeros are in the number 5! (5 factorial)? 9 AnswersThere are 24 trailing zeros. 100*99*98*97*...*2*1. A simple brain-teaser meant to recognize patters of multiplying successive multiples of ten and five. 5!=120. So there is 1 trailing zero. 5!=120. So there is 1 trailing zero. Show More Responses This sounds like one geared not so much towards getting the right answer, but getting to it the right way. If you think a bit and say "one", the interviewer will know you did it the brute-force way, doing the math. You'd get at the answer faster, and probably impress them more, if you think instead how many times a ten will be produced in doing that math, rather than what the actual result of the math will be. Sorry, I came back here to find what this BS question was and saw that I mistyped it. How may trailing zeros in (100!) ? Can someone explain to me exactly what does this question have to do with systems engineering? Do they present the question to weed out non-academics? I never met an academic who was an effective engineer. >I never met an academic who was an effective engineer. Really? Your logic seems to show how ineffective of an engineer you are. If Google hires based on this practice, and Google is the de-facto leader in engineering, then what they are doing is the correct solution. http://www.purplemath.com/modules/factzero.htm To generate a zero, we need a (5,2) factor pair. For any given number N, we have at least N/2 number of multiples 2, so the number of zeroes can be determined by the count of numbers that have 5 as a factor (i.e we have more 2s than 5s) Roughly, we can count N/5 numbers that are multiples of 5, add to that numbers that are multiples of 5^2 (these will have two 5 factors) i.e N/25, add to that numbers that are multiples of 5^3 (these will have three 5 factors) and so on. For eg: 10! -> 2 multiples of 5 -> 2 zeroes 100! -> 20 multiples of 5 + 4 multiples of 25 -> 24 zeros 500! -> 100 multiples of 5 + 20 multiples of 25 + 4 multiples of 125 -> 124 zeros 1000! -> 200 multiples of 5 + 40 multiples of 25 + 8 multiples of 125 (5^3) + 1 multiple of 625 (5^4) -> 249 zeros |

### Program Manager at Microsoft was asked...

You are on a game show. There are three doors, behind one of which is a prize and the other two is a chunk of coal, and the host knows which door holds the prize. You choose door #1. Before it is opened, the host opens door #3 and reveals a lump of coal. You have the choice to stick with the door you chose originally or switch to door #2. What do you do? 17 AnswersSwitch doors. When you chose door #1, there was a 66% chance that the prize was not behind that door. When the host revealed the coal, there was still a 66% chance the prize was not behind the door you chose. Thus, you have double the odds of getting the prize by switching to door #2. The key to this puzzle is that the host knew which door has the prize. That is not true. The probability went from 1/3 to 1/2 once the number of doors reduced to 2. However, statistically speaking, your chances of finding the treasure are now even. So it should not matter which door you pick. Interview candidate is right. You got 1/3 chance that prize is behind door #1 and you lose if you switch. And you got 2/3 chance that prize is behind either door #2 or #3. Since the host will always eliminate the wrong one. 2/3 chance will be allocated on the left one. Show More Responses Everyone is wrong. You stay with your first pick, the odds becoming slanted greatly against the other door. Interview Candidate and Anonymous are right. This is also known as the Monte Hall problem Your choice splits the doors in two sets. Set A contains the door you selected, and the probability that is a prize behind this door is 1/3. The set B contains all remaining doors, and the probability that the winning door is somewhere in there is 2/3. By removing one door, which all have the success probability of zero because there's coal behind them, from set B, only one door remains in B, but the overall probability for success in set B is still 2/3. Therefore you must switch. I can't believe that some of the answers up here actually state that you should switch. -------------------------- Before the host showed you that door#3 was hiding coal, your chances to pick the right door where 1/3. Now that you know that coal is behind door#3, you only know that 1 of the 2 remaining doors leads to coal while the other leads to the prize. While your probability of making the right choice has increased from 1/3 to 1/2, this probability still applies to both of the doors. You can't differentiate them. Seriously. Tarik, You are correct that the right choice has increased from 1/3 to 1/2, but you have to account for variable change in this instance. Think of the same problem, but with different numbers. 100 doors, same conditions. You pick 1 door, with a 1/100 chance of being correct. The game show opens 98 doors with coal (because he knows where the prize is). He now offers you the chance to keep your door or switch to the last remaining door. In this Because of variable change you will always be better off switching doors. Scratch that, the right choice hadn't increased from 1/3 to 1/2. It's late and I have no idea why I typed that. Really??? I can see why they ask the question if so many people have so little understanding of probability. There are NOW two doors to choose from and there is one prize. DAAAAA It doesn't matter how many there were to begin with or which one of the two you originally chose. There are now two doors and each one has an equal chance of containing the prize so your odds of success do not change at all regardless if you stay with your original choice of switch to the other. The odds of your original door being the correct one were 1/3 when there were 3 option but the odds of your door being the correct one CHANGE to 1/2 when you eliminate one of the options. To say the probability of your door being correct are still 1/3 is to say that your odds of success are still one third if you were to pick door 3 even though you know it does not have the prize. As an interviewer I could have easily eliminated the majority of the applicants above. What about sticking to your original decision and not deviating without knowing the facts(Data). i will stick with my decision and might fail, sure but at list i will not make decision on just the 1/3 and 1/2 number game. Averee has given the best explanation. Before anyone writes any further comment or contradicts what he says, do yourself a favor, go on youtube and watch the video on monty hall problem - you will not object to averee's solution after that. What everyone who is answering "stick to original door" is missing the host's knowledge of which door has the prize and his actions after the first door is chosen. Host's action puts the odds of prize behind the switched door as 0.5. However, the original door still has a 0.33 probability of the prize. Yes I know it is a difficult concept to grasp - so you all need to watch the monty hall problem on youtube. Show More Responses Switching is the correct answer. Thank you Akhil. I'd like to add that the youtube video "Monty Hall Problem for Dummies - Numberphile" explained it very well. PS. Watch it twice. You must switch the door. Imagine the same situation with 100 doors and you will understand. |

You are in a room with 3 switches which correspond to 3 bulbs in another room and you don't know which switch corresponds to which bulb. You can only enter the room with the bulbs once. You can NOT use any external equipment (power supplies, resistors, etc.). How do you find out which bulb corresponds to which switch? 6 AnswersTurn on 2 switches. Turn one switch off after a few minutes. Enter room with bulbs. The bulb that is on is the only switch that is on. The bulb that is off but hot, is the switch that has been turned off. The bulb that is off and cold is the third switch. Turn on switch A for 1 minute. Turn it off. Turn on switch B. Enter the room with the bulbs. The hot bulb is A The lit bulb is B The dark bulb is C Turn switch A leave it on for 1/2 minutes and turn it off then turn on switch B. Now goto the room, the one Hot is A, Cold(or not Hot) is C, the bulb thats on is B. Show More Responses The answers here all have two assumptions which are unwarranted. 1. Why do you think the light emits heat? 2. Why do you think you can touch the bulbs? (3. If you're never going to enter the room again what's the point of knowing which switch goes with which light?) The best you can do is reduce to guessing one of two bulbs. RE: Objection 1. I guess you've never touched the outside of a light bulb after it's been on for awhile. The filament gets hot-it glows, the light we see is that glowing-the heat from from the glowing filament transfers to the glass-the glass is hot. 2. Why not? This is an exercise to demonstrate problem-solving ability, not a "MacGyver" episode. Who's to say the candidates didn't ask about the room's dimensions or where the bulbs were located, or that assumptions could be made in order to formulate an answer? 3. I think you missed the point of the question. 4. Sheesh. 3. Yes, Objection, I did see that I entered "3." on here a second time. BTW, I'm not returning to this forum again, so any reply/retort/threat won't be seen by me. Good luck in your job hunting. |

### Inventory Specialist at Apple was asked...

What was the hardest thing you ever had to do in your prior work experience? 7 AnswersLeave Climb out on a ledge in order to weld a beam no harness 50 feet up, nothing to hold onto once my welding helmet goes down and i lean over. I got out there and couldn't make myself let go. I was walking on a 3" wide beam. I failed. my scared ass had to get another idiot to do it. lol I worked in starting based organisation. There were v.less seniors. So, if there is any issue, I alone did R&D and resolved it. Though, I took time to resolved, but that made me aware of many scenerios and gave confident boost. Show More Responses Travel to Hawaii to fire a rep. It was bad because there was no time to stay and relax. The owner insisted that I return immediately! Had to terminate a female employee who was badly injured from a fire. It had been proven via testimony to us that she covered herself with gasoline and lit it off to get even with a boyfriend, but she said it was due to an accidental kitchen fire. Fire Marshall report also verified she was not telling truth. Our policy said no coverage from 'self-inflicted' incidents and she had lied to us about the cause of the fire. As I told her she was being terminated, I was looking at a melted face, and compresses on each of her arms, and in a torso cast. This was three months post incident and investigation. You don't feel good after a day like that one! This is a comment. Steve... wow. THAT'S not hard... that's unreal. The hardest thing I had to do is doing something for others |

You have 1 seat left on a flight, and you have 5 passengers waiting on standby, a military man in uniform, a pregnant woman, a woman and her infant child, an elite customer(one who is a frequent flyer of NWA), and a gentleman trying to go an see his ill sister. Who will be the one to get the empty seat on the flight, and why? 8 AnswersUnless it is life or death, our Elite customers always have preference, they are frequent flyers and therefor they are providing alot of revenue for the airline. Not the pregnant woman, if she may be in fear of labour, it would be best she delay her flight not hurry-up and get on. Not the woman with infant because there is no priority here except being polite perhaps. Same with the man visiting his ill sister (didn't say it was life/death situation) It is true that NWA gives preference to the frequent flyers and would normally be the one to get the seat HOWEVER the Military man, IF he is on active duty and believes there is a threat to the aircraft and/or may involve National Security (which may in-effect mean life or death) would get the seat- I believe that post-911 this supersedes ALL normal procedures at any US airport or may or may now be included in the new procedures. The first one to get on the standby list. Show More Responses Isn't this the decision for the boarding gate agent, not the flight attendant. it's my first try to assist to a pregnant women in getting a seat in first. It would be the man who is a frequent flyer I would give it to the military man on the basis that his service is a part of the reason we get to fly back and forth whenever we want and need to freely. The first thing it is the gate attendants decision not the flight attendants. But the one who should get the seat is the passenger who has the family member that is ill. That family member may be terminal and that passenger may need to get to his destination as soon as possible. |