# Data engineer intern Interview Questions

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Data Engineer Intern interview questions shared by candidates### Suppose you had eight identical balls. One of them is slightly heavier and you are given a balance scale . What's the fewest number of times you have to use the scale to find the heavier ball?

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

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

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

### Write an SQL query that makes recommendations using the pages that your friends liked. Assume you have two tables: a two-column table of users and their friends, and a two-column table of users and the pages they liked. It should not recommend pages you already like.

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CREATE temporary table likes ( userid int not null, pageid int not null ) CREATE temporary table friends ( userid int not null, friendid int not null ) insert into likes VALUES (1, 101), (1, 201), (2, 201), (2, 301); insert into friends VALUES (1, 2); select f.userid, l.pageid from friends f join likes l ON l.userid = f.friendid LEFT JOIN likes r ON (r.userid = f.userid AND r.pageid = l.pageid) where r.pageid IS NULL; Less

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select w.userid, w.pageid from ( select f.userid, l.pageid from rollups_new.friends f join rollups_new.likes l ON l.userid = f.friendid) w left join rollups_new.likes l on w.userid=l.userid and w.pageid=l.pageid where l.pageid is null Less

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SELECT f.userid, l.pageid FROM friends f LEFT JOIN likes l ON f.friendid = l.userid WHERE l.pageid NOT IN (SELECT r.pageid FROM likes r WHERE f.userid = r.userid) Can someone tell me if this works? Less

### 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?

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Bayesian 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 ;) Less

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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. Less

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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. Less

### want you to write me a simple spell checking engine. The query language is a very simple regular expression-like language, with one special character: . (the dot character), which means EXACTLY ONE character (it can be any character). So, for example, 'c.t' would match 'cat' as the dot matches any character. There may be any number of dot characters in the query (or none). Your spell checker will have to be optimized for speed, so you will have to write it in the required way. There would be a one-time setUp() function that does any pre-processing you require, and then there will be an isMatch() function that should run as fast as possible, utilizing that pre-processing. There are some examples below, feel free to ask for clarification. Word List: [cat, bat, rat, drat, dart, drab] Queries: cat -> true c.t -> true .at -> true ..t -> true d..t -> true dr.. -> true ... -> true .... -> true ..... -> false h.t -> false c. -> false */ // write a function // Struct setup(List<String> list_of_words) // Do whatever processing you want here // with reasonable efficiency. // Return whatever data structures you want. // This function will only run once // write a function // bool isMatch(Struct struct, String query) // Returns whether the query is a match in the // dictionary (True/False) // Should be optimized for speed

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bear in mind for your solution, checking the lengths of words in the dictionary is very fast. That's what you can use your setup for. There's no need to iterate through the whole loop of checks if the word fails the length already. See my solution above Less

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Here is the Python Code (inspired by someone's code on this page): def setUp(word, input_list): word = word.strip() temp_list = [] Ismatch = False if word in input_list: Ismatch = True elif word is None or len(word) == 0: Ismatch = False else: for w in input_list: if len(w) == len(word): temp_list.append(w) for j in range(len(temp_list)): count=0 for i in range(len(word)): if word[i] == temp_list[j][i] or word[i] == '.': count += 1 else: break if count == len(word): Ismatch = True print(Ismatch) def isMatch(word, input_list): return setUp(word, input_list) isMatch('c.t', ['cat', 'bte', 'art', 'drat', 'dart', 'drab']) Less

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def spellchecker(word_list,spell): # if spell in word_list: # return True k=False for word in word_list: if len(spell)==len(word): k= all(spell[i]==word[i] or spell[i]=='.' for i in range(len(spell))) return k w=['cat', 'bat', 'rat', 'drat', 'dart', 'drab'] r='d.a.' print(spellchecker(w,r)) Less

### Write a SQL query to compute a frequency table of a certain attribute involving two joins. What if you want to GROUP or ORDER BY some attribute? What changes would you need to make? How would you account for NULLs?

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select number_comments, count(submission_id) as number_posts from ( # more than zero comments select submission_id, count(post_id) as number_comments from ( select submission_id, case when parent_id is null 1 else 0 end as post, case when parent_id is not null parent_id else null end as post_id, body from Submissions )k where post =0 group by submission_id ) k1 group by number_comments union select number_comments, count(submission_id) as number_posts from ( # comments= 0 select submission_id, 0 as number_comments from ( select submission_id, case when parent_id is null 1 else 0 end as post, case when parent_id is not null parent_id else null end as post_id, body from Submissions )k where post =1 group by submission_id ) k1 group by number_comments Less

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I've tested all these on a mock data set and none of them work! Does anyone have the correct solution? I'm stuck on this one.. Less

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Posts and comments in the same table looks weird. Here's my attempt (made easy with CASE) to exclude all the posts from the table and grouping/counting comments. SEL parent_id ,COUNT(*) as comment_count ( SEL * ,CASE WHEN perent_id IS NULL THEN 'Post' ELSE 'comment' END as post_or_comment FROM Submissions ) a WHERE post_or_comment = 'comment' Less

### Given an list A of objects and another list B which is identical to A except that one element is removed, find that removed element.

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All these supposed answers are missing the point, and this question isn't even worded correctly. It should be lists of NUMBERS, not "objects". Anyway, the question is asking how you figure out the number that is missing from list B, which is identical to list A except one number is missing. Before getting into the coding, think about it logically - how would you find this? The answer of course is to sum all the numbers in A, sum all the numbers in B, subtract the sum of B from the sum of A, and that gives you the number. Less

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select b.element from b left join a on b.element = a.element where a.element is null Less

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In Python: (just numbers) def rem_elem_num(listA,listB): sumA = 0 sumB = 0 for i in listA: sumA += i for j in listB: sumB += j return sumA-sumB (general) def rem_elem(listA, listB): dictB = {} for j in listB: dictB[j] = None for i in listA: if i not in dictB: return i Less

### Given an array of integers, we would like to determine whether the array is monotonic (non-decreasing/non-increasing) or not. Examples: // 1 2 5 5 8 // true // 9 4 4 2 2 // true // 1 4 6 3 // false //1 1 1 1 1 1 // true

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if l == sorted(l) or l == sorted(l,reverse=True): print(True) else: print(False) Less

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python solution. inc = dec = True for i in range(len(nums)-1): if nums[i] > nums[i+1]: inc = false if nums[i] < nums[i+1]: dec = false return inc or dec Less

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An array is monotonic if and only if it is monotone increasing, or monotone decreasing. Since p = A[i+1] for all i indexing from 0 to len(A)-2. Note: Array with single element can be considered to be both monotonic increasing or decreasing, hence returns “True“. --------------------Python 3 code---------------------- # Check if given array is Monotonic def isMonotonic(A): return (all(A[i] = A[i + 1] for i in range(len(A) - 1))) # Test with an Array A = [6, 5, 4, 4] # Print required result print(isMonotonic(A)) Less

### Data challenge was very similar to the ads analysis challenge on the book the collection of data science takehome challenge, so that was easy (if you have done your homework). SQL was: you have a table where you have date, user_id, song_id and count. It shows at the end of each day how many times in her history a user has listened to a given song. So count is cumulative sum. You have to update this on a daily basis based on a second table that records in real time when a user listens to a given song. Basically, at the end of each day, you go to this second table and pull a count of each user/song combination and then add this count to the first table that has the lifetime count. If it is the first time a user has listened to a given song, you won't have this pair in the lifetime table, so you have to create the pair there and then add the count of the last day. Onsite: lots of ads related and machine learning questions. How to build an ad model, how to test it, describe a model. I didn't do well in some of these.

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Can't tell you the solution of the ads analysis challenge. I would recommend getting in touch with the book author though. It was really useful to prep for all these interviews. SQL is a full outer join between life time count and last day count and then sum the two. Less

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Can you post here your solution for the ads analysis from the takehome challenge book. I also bought the book and was interested in comparing the solutions. Also can you post here how you solved the SQL question? Less

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for the SQL, I think both should work. Outer join between lifetime count and new day count and then sum columns replacing NULLs with 0, or union all between those two, group by and then sum. Less

### We have two options for serving ads within Newsfeed: 1 - out of every 25 stories, one will be an ad 2 - every story has a 4% chance of being an ad For each option, what is the expected number of ads shown in 100 news stories? If we go with option 2, what is the chance a user will be shown only a single ad in 100 stories? What about no ads at all?

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For the questions 1: I think both options have the same expected value of 4 For the question 2: Use binomial distribution function. So basically, for one case to happen, you will use this function p(one case) = (0.96)^99*(0.04)^1 In total, there are 100 positions for the ad. 100 * p(one case) = 7.03% Less

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For "MockInterview dot co": The binomial part is correct but you argue that the expected value for option 2 is not 4 but this is false. In both cases E(x) = np = 100*(4/100) = 4 and E(x) = np=100*(1/25) = 4 again. Less

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Chance of getting exactly one add is ~7% As the formula is (NK) (0,04)^K * (0,96)^(N−K) where the first (NK) is the combination number N over K Less

### Consider a game with 2 players, A and B. Player A has 8 stones, player B has 6. Game proceeds as follows. First, A rolls a fair 6-sided die, and the number on the die determines how many stones A takes over from B. Next, B rolls the same die, and the exact same thing happens in reverse. This concludes the round. Whoever has more stones at the end of the round wins and the game is over. If players end up with equal # of stones at the end of the round, it is a tie and another round ensues. What is the probability that B wins in 1, 2, ..., n rounds?

16 Answers↳

Because at the beginning time, A has 8 and B has 6, so let A:x and B:y, then A:8+x-y and B:6-x+y; so there are 10/36 prob of B wins. And A wins prob is 21/36 and the equal prob for next round is 5/36. So for B wins at round prob is 10/36. And if they are equal and to have another round, the number has changed to 7 and 7. So A:7+x-y and B:7-x+y, so this time B wins has prob 15/36 and A wins has prob 15/36. And the equal to have another round is 6/36=1/6. So overall B wins in 2 rounds has prob 5/36*15/36. And for round 3,4,...etc, since after each equal round, the number will go back to 7 and 7 so the prob will not change. So B wins in round 3,4,...n has prob 5/36*(6/36)^(r-2)*15/36. r means the number of the total rounds. Less

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So many answers...Here's my version: For round1, B win only if it gets 3 or more stones than A, which is (A,B) = (1,4) (1,5) (1, 6) (2, 5) (2,6) (3,6) which is 6 cases out of all 36 probabilities. So B has 1/6 chance to win. To draw, B needs to get exactly 2 stones more than A, which is (A, B) = (1,3) (2,4) (3,5) (4,6) or 1/9. Entering the second round, all stones should be equal, so the chance to draw become 1/6, and the chance for either to win is 5/12. So the final answer is (1/6, 1/9*5/12, (1/9)^2*5/12, .....(1/9)^(n-1)*5/12) ) Less

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I don't get it. Shouldn't prob of B winning given it's tie at 1st round be 15/36? given it's tie at 1st round, at the 2nd round Nb > Na can happen if (B,A) is (2,1), (3,1/2),(4,1/2/3), (5,1/2/3/4),(6,1/2/3/4/5), which totals 15 out of 36. Less