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When they ask a quesion like this at MS, they do want an answer. If you tell them that you want to consider alternatives up front, they will wave that off and tell you that, in this hypothetical situation, alternatives were already considered and that moving the mountain is the approach was chosen. They really want you to answer the question. The point of this question is - process. They want to see what process you use to solve problems. It is important to show that you solve the problem not by arranging and re-arranging a series of random thoughts but that you can approach it methodically and that this methodology can be applied to any problem. Do not to to some up with a clever answer that attempts to solve the problem - they will just keep insisting that you tackle the problem. If you don't, you won't pass the interview. So, brush up on your problem solving process before you interview at MS. Use these questions as an opportunity to impress them with how well you can solve difficult problems. Menos

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This is a common dorky Computer Science joke. The answer I believe they are looking for is that you use the Tower of Hanoi algorithm to move the mountain (i.e. that the problem of moving Mt. Fuji is reducible to the already-solved Tower of Hanoi problem). This could be accomplished by having a large laser and a couple of really good cranes. Menos

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This is a straightforward programming problem. Clever solutions such as allowing the other trucks to refuel and allowing other trucks to tow extra trucks full of fuel rarely impress the interviewer. The simplest solution involves recursion, or induction. Imagine a function f(n) where f(n) is the distance n trucks can carry the load, the problem defines f(1)=100, and we're asked to solve for f(50), so our job is to solve for f(n) in terms of f(n-1), f(1) and n. It's safe to assume each truck, including the fully loaded truck will travel the same distance and that fuel is used uniformly over the distance. So with n trucks, the trucks should travel just far enough that n-1 trucks have room in their tanks for the nth truck's fuel, then transfer an solve for n-1. This equation is f(n) = f(1)/(n) + f(n-1) (All perl vs python vs ruby/etc wars aside) Plugging this into a simple scripting language such as perl is an easy way to solve this: sub f{ my $n = shift; return 100 if($n == 1); return f(1)/($n) + f($n-1); } print "50: " . f(50) . "\n";' On the command line, this gives a quick answer: > perl -e 'sub f{ my $n = shift; return 100 if($n==1); return f(1)/($n) + f($n-1);} print "50: " . f(50) . "\n";' 50: 449.920533832942 Approximately 449.92 miles with 50 trucks. The recursive subroutine above will suffice as a general solution in terms of n. Menos

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So 50 trucks is solvable but annoying. The answer depends on payload size, also if the truck needs to return. If we assume no return and 1 truck can carry the payload. Make the number of trucks 64 for even powers of 2. The trucks all take off together and every 50 miles they all have half full tanks. Half of the trucks are sacrificed to refill the other half. Number of trucks goes from 64 -> 32 -> 16 -> 8 ->4 ->2 -> 1. So 6 splits * 50 miles = 300 + last truck has 100 miles. 400 for 64 trucks. 50 can be solved with annoying fractions. General answer is 50 * log(base2) of n + 100. :-) Menos

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I took the test in c++. The function interfaces were in base c, though, so it would probably look more or less identical in c or java. Menos

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Yeah, it was for me. There was few days pause between the 2nd round and the offer, so maybe they looked at other parts of my resume and decided to forgo a 3rd round for me. So I really can't say for sure that you won't have more rounds. Menos

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Hey, I just have one question out of curiosity, was the second online assessment web-cam proctored? Menos

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i heard amazon is changing their recruiting process. Many people complained that the online proctoring was an invasion of their privacy. I also had a second online assessment but it was not webcam proctored. I am assuming the recruiting process is now two online non webcam assessments then a final phone interview. My friend had that round of interviews with Amazon 2 weeks ago. Menos

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Hi, did you interview on Jan 27? Did the engineer ask about behavioral questions or resume questions or did they just jump right into the coding questions? Also I finished my second assessment a week ago but still didn't get a reply back. Was that normal for you? Menos

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Hi, did you interview on Jan 27? Did the engineer ask about behavioral questions or resume questions or did they just jump right into the coding questions? Also I finished my second assessment a week ago but still didn't get a reply back. Was that normal for you? Menos

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The recruiter called me yesterday and informed me that they are no longer gonna recruit new college grads for SDE positions because they already have enough. She told that they are done with 2017 and 2018 recruitment. This was through University Programs. I was invited for an onsite and then revoked. It was a little annoying because they waited 2 weeks to inform me this (that too on a Friday eve!) I spent the 2 weeks preparing for this one, when I could have given Twitter. But again, I am nothing and they are people of power, so what I feel doesn't matter! Menos

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What a shame! So basically, whoever hasn't been responded back to would mostly be rejected. Hence no point waiting!! Menos

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``` // Lexically nested function definitions defined within enclosing function function Sum(arg0) { function Inner1(arg1) { function Inner2(arg2) { return arg0 + arg1 + arg2; } return Inner2; } return Inner1; } console.log(Sum(3)(4)(5)); ``` Menos

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var sum=0; // global variable function Sum(num){ sum=num+sum; return Sum; } Sum(3)(4)(5); Menos

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Easiest way to do it. Written in python. If you consider the easiest case, when our summed value (k) is 0, the pairs will look like -50 + 50 -49 + 49 -48 + 48 etc.... etc... So what I do is generalize the situation to be able to shift this k value around. I also allow us to change our minimums and maximums. This solution assumes pairs are commutative, i.e. (2, 3) is the same as (3, 2). Once you have the boundaries that you need to work with, you just march in towards k / 2. This solution runs in O(n) time. def pairs(k, minimum, maximum): if k >= 0: x = maximum y = k - maximum else: x = k + maximum y = minimum while x >= k / 2 and y <= k / 2: print str(x) + " , " + str(y) + " = " + str(x + y) x = x - 1 y = y + 1 Menos

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here is my solution using hash table that runs in O(2n) => O(n): public static String findNums(int[] array, int sum){ String nums = "test"; Hashtable lookup = new Hashtable(); for(int i = 0; i < array.length; i++){ try{ lookup.put(array[i], i); } catch (NullPointerException e) { System.out.println("Unable to input data in Hashtable: " + e.getMessage()); } } int num2; int num1; for (int i = 0; i < array.length; i++){ num2 = sum - array[i]; Integer index = (Integer)lookup.get(num2); if ((lookup.containsKey(num2)) && (index != i)){ num1 = array[i]; nums = array[i] + ", and " + num2; return nums; } } //System.out.println(lookup.get(-51)); return "No numbers exist"; } Menos

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@Ja, Your example doesn't really make sense: how can the node with value 6 point to a node with value 7 AND a node with value 8? It can only point to one node: either 7 or 8. That's why I think Anonymous' answer is correct. Menos

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1) len1=find length of linkedlist1 2) len2 =find length of linkedlist2. 3) move the bigger linked list to (len1-len2) position. 4) rightnow both linked lists are equal at distance from last node. that is they are n node away last node. 5) iterate both LL simulatenously and if they have same instance that is their intersection point. Menos