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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 Minder

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Use Except select f.user_id, l.page_id from friends f inner join likes l on f.fd_id = l.user_id group by f.user_id, l.page_id -- for each user, the unique pages that liked by their friends Except select user_id, page_id from likes Minder

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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. Minder

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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. Minder

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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 Minder

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@ RLeung shouldn't you use left join? You are effectively losing all posts with zero comment. Minder

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

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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 Minder

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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? Minder

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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. Minder

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Patience, though it is a virtue, it can be a gift when you try to understand a co-worker rather than gossip, yell or ostracize them. Minder

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Time. I spent an afternoon in an assisted-living home and helped with the Bingo tournament. It was all the rage and my ounce of time is immeasurable to those who need it the most. Minder

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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. Minder

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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 Minder

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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) ) Minder

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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. Minder

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find the right most element. If this is a right node with no children, return its parent. if this is not, return the largest element of its left child. Minder

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One addition is the situation where the tree has no right branch (root is largest). In this special case, it does not have a parent. So it's better to keep track of parent and current pointers, if different, the original method by the candidate works well, if the same (which means the root situation), find the largest of its left branch. Minder