# Preguntas de entrevista de Development engineer

# 25 mil

Preguntas de entrevista para Development Engineer compartidas por los candidatos## Principales preguntas de entrevista

### Given a string "aaabbbcc", compress it, = "a3b3c2" . Given that output string's length is always smaller than input string, you have do it inplace. No extra space

29 respuestas↳

#include #include #include #define STR_SIZE 26 int main() { /* Current char sequence tracker */ char *c = NULL; char *b = NULL; char *str = (char*)malloc(STR_SIZE * sizeof(char)); if(NULL == str) return -1; memcpy(str, "aaaabbbcceeeeefffffff", 26); b = c = str; printf("Input: %s\n", b); while(*str) { if(*(str+1) != *str) // Repeat sequence ends { // Add 48 so the count gets printed as a char *(c+1) = ((str-c+1)+48); // Updated count, copy rest of the string starting after the count position memcpy(c+2, str+1, strlen(str)); // Update c to point to the new char repeat sequence c = c+2; } str++; } /* b was initialized to point to str up top, proving it was done in place */ printf("Output:%s\n", b); free(b); b = NULL; return 0; } Menos

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Using Ruby methods, ans = "" s.split("").group_by{|x| x}.each{|k,v| ans += "#{k}#{v.count}"} puts ans But this code snippet will produce "a4b3c2" for the input string "aaabbbcca" Menos

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string CombineWithNoExtraSpace(string S) { int n=S.size(),count=1,i=0; if(i==n) return "\0"; while(i+1<=n && S[i]==S[i+1] ) { count++; i++; } S = S[i] + to_string(count) + combine(S.substr(i+1,n-count)); return S; } Menos

### how would you move mount fuji?

27 respuestas↳

I would first answer with, "First, I would analyze the problem and determine if it didn't make better sense to come to the mountain rather than move the mountain. Assuming that's not feasible..." I think that's a key element they're looking for in an answer. That you can look at a major task and first identify if there isn't a better approach. The next element is to determine how you would go about completing a seemingly impossible or gargantuan task. The specifics of this part of the answer don't matter other than to show that you have an understanding that huge problems need to be broken down into smaller, more manageable tasks using the resources you have. Menos

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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

### Given a fleet of 50 trucks, each with a full fuel tank and a range of 100 miles, how far can you deliver a payload? You can transfer the payload from truck to truck, and you can transfer fuel from truck to truck. Extend your answer for n trucks.

23 respuestas↳

The answer is 450 miles. it is the sum for n=1 to 50 of [ 100 / (50-n) ] After 2 miles you would transfer all of the fuel from one truck, and could fill all 49 of the other trucks. (100mile range / 50 trucks = 2 miles before the first truck could be emptied of all fuel and is not needed to carry any fuel) at that point you only have 49 trucks, so 100/49 is the distance before you park the second truck. After parking 49 trucks you would have one full truck of fuel that would finish the last 100 miles. General solution for X trucks with Y range is Sum for n=1toX of [Y / (x-n)] 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

### non disclosure agreement

18 respuestas↳

I don't actually know. I mean, that's what I would do if I were amazon, but I did not read anything about them webcamming me, and I also didn't happen to notice the webcam light being on. That didn't mean it didn't happen, though. 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

### What was one of your best achievements on a project in the past?

18 respuestas↳

From what i've heard and seen, just wait, following up will barely do you any good. 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

### DS, Algorithms.

18 respuestas↳

Review Leetcode, CTCI.

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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

### Discussed online assessment.

17 respuestas↳

Yes i did tried to contact recruiter.. no response back. Did you get any response ? 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

### Implement Sum(3)(4)(5)=12 with javascript

17 respuestas↳

const Sum = a => b => c => a + b + c;

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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

### Given a set of numbers -50 to 50, find all pairs that add up to a certain sum that is passed in. What's the O notation for what you just wrote? Can you make it faster? Can you find an O(n) solution? Implement the O(n) solution

16 respuestas↳

Put all the numbers from the array into a hash. So, keys will be the number and values of the keys be (sum-key). This will take one pass. O(n). Now, foreach key 'k', with value 'v': if k == v: there is a match and that is your pair. this will take another O(n) pass totale O(2n) ~ O(n) 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

### Most of them were expected. Almost all are problem solving questions. 1. Given a BST with following property find the LCA of two given nodes. Property : All children has information about their parents but the parents do not have information about their children nodes. Constraint - no additional space can be used

15 respuestas↳

function findLCA(Node node1, Node node2) { int counter1 = 0; int counter2 = 0; Node temp; //Find the level for each node, use a temp node to //traverse so that we don't lose the info for node 1 and node 2 temp = node1; while( temp.parent ! = null) { temp = temp.parent; counter1++; } temp = node2; while( node2.parent ! = null) { node2 = node2.parent; counter2++; } /* * We wanna make them at the same level first */ if(counter1 > counter2) { while(counter1 != counter2) { node1 = node1.parent; counter1--; } } else { while(counter2 != counter1) { node2 = node2.parent; counter2--; } } while (node1.parent != node2.parent) { node1 = node1.parent; node2 = node2.parent; } System.out.println("Found the LCA: " + node1.parent.info); } Menos

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//correction temp = node2; while( temp.parent ! = null) { temp = temp.parent; counter2++; } Menos

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Consider this is a BST, where max node is always on the right of min node, we can traverse max upward one node at a time while comparing min nodes as it traverse upward toward root. BinaryNode findBSTLCA( BinaryNode min, BinaryNode max ) { BinaryNode tempMax = max; BinaryNode tempMin = min; while( tempMax != null ) { while( tempMin != null ) { if( tempMin.element == tempMax.element ) return tempMin; tempMin = tempMin.parent; } tempMin = min; // reset tempMin tempMax = tempMax.parent; // traverse tempMax upward 1 node } return null; // no LCA found } Menos