Intern Developer Interview Questions | Glassdoor

# Intern Developer Interview Questions

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Intern developer interview questions shared by candidates

## Top Interview Questions

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Aug 13, 2013

Feb 15, 2012

### Financial Software Developer Intern at Bloomberg L.P. was asked...

Apr 17, 2012
 There are 20 floors in a building. If you're on an elevator and you're trying to get to the 20th floor, what is the probability that 4 people ahead of you click the 20th floor before you do? Assuming you click last.10 Answersassume there is one button for each floor, so 20 buttons. a person can press any 1 button out of the 20, prob is 1/20. Since there are 4 people, so1/16000These are independent events so the chances of one person before you going to the 20th floor is 1/20. Since this happens 4 times before you the probability is 4*(1/20) or 1/5.The above two are close, but wrong.. There are 20 buttons, thus 20 choices, sure. But you are getting on at one of the floor. No body will press the button for the floor they get on.. Thus, there is really only 19 choices. P = (1/19)^3 (Independent events mean (1/19)(1/19)(1/19)).Show More Responses1/19 + 1/18 + 1/17 + 1/16 assuming that there were no repeated destinations.based on question: P(all 4 ahead of you want to get off on 20th fl) = (1/19)^4 real life(all 4 want to get off on 20th fl, and one of them is the first person press the button to 20th fl, and that leave all others, including you, stay still): (1/19) * (1/4)about 20% is the right answer. I am surprised with some of the answers, they are all very small possibilities (some less than 1%).I'm quite sure you are all wrong: The real probability is 1 - P(nobody pushes 20) = 1 - (18/19)^3 = 15%1- (19/20)^4If one of the 4 press the button for the 20th floor then the others won't have to do anything. The chances of one of them pressing 20th is: 1/19 + 1/19 + 1/19 + 1/19 = 4/19The answer is 1-(19/20)^4

### Software Development Engineering Intern at Amazon was asked...

Jun 23, 2012
 You are given an array with n positive integers where all values in the array are repeated except for one. Return the one that is not repeated.7 Answerspublic static int notRepeated(int[] given) { Map m = new HashMap(); for(i=0; i < given.length; i++) { if(m.get(given[i])) { m.put(given[i], 2); } else m.put(given[i], 1); for(x:m) { if(x == 1) return x; } } }If you have an array of positive integers with only ONE non repeated integer, the easiest thing is just xor them all. Whatever you return will be the answer. Perl answer below sub findNotRepeated{ my (\$arr) = @_; if(@\$arr==0) { return - 1; } my \$val = 0; foreach(@\$arr) { \$val ^= \$_; } return(\$val); } sub findMissingElement{ my (\$arr,\$arr2) = @_; if(@\$arr==0 || @\$arr2 == 0 ) { print " arr2=" .@\$arr2 . "X\n";; return - 1; } my \$val = 0; foreach((@\$arr , @\$arr2)) { \$val ^= \$_; } return(\$val); }first sort the array.n then 1)for i=1 to n { if ( element [i] !=element [i+1] ) { cout<

### Software Development Engineer Intern at Amazon was asked...

Feb 15, 2012
 To return the 'm' smallest numbers from a file of 'n' numbers8 AnswersI would first create an array holding the first m values of the file, and sort. Then, each value I read from the file, I can compare the the last (largest) value in the array. If its smaller, do a binary search to find the correct spot in the array. Insert the number and shift everything behind it back by 1. Worst case runtime will be O(n lg m)Why cant we simply start with min=INT_MIN. Then, read first number from file, if lower, set min=that number. Seek next number and so on... We dont need to load the whole file. We will go to the disk often though.I will create a min heap with the given numbers and extract 'm' minimums from the heap which will get stored into the resultant arrayShow More ResponsesAnuj, Wont it take time of O((M+N)log N)@spammer, I think it's O(n+m*log n), since you can construct the min heap bottom-up and that only takes O(n)Why don't we just sort the array and return the first m numbers? Takes O(nlogn) for sorting and O(m) to return the first m numbers.Modified quick sort. Pick a pivot and test if one half is smaller than m. If it is, drag the rest from the other half; if it is not, keep partitioningMaintain a heap of m numbers and also track a variable which stores the current highest value. So as u go through the n numbers, if the number is higher than the current highest number, ignore it else insert it into the heap and set the current highest value to the newly inserted value

### Software Development Engineer Intern at Amazon was asked...

Jan 6, 2011
 Write a program to find the square root of a double.5 Answersuse binary search to narrow down the search space and keep multiplying the answer in each iteration by itself to check if it is equal to the double given. If it is lesser, move up the lower bound, else move down the upper bound.One easy to remember method that also has much better performance than binary searching for the result is the Babylonian Method. It is similar to Newton's method for finding derivatives. See the algorithm here: http://en.wikipedia.org/wiki/Methods_of_computing_square_roots#Babylonian_method Also, combining this algorithm with the Rough Estimation also described on that page will compute the squareroot of most double numbers in less than 10 iterations.Is it too obvious to ask if you can do double^.5 ?Show More ResponsesI would respond by showing that I am thinking about the problem as it is defined, not by making assumtions about what is implied. This can result in product that does not meet the requirement specifications, which can be very costly: "What do you mean, a program or a function? A program would require input and output mechanisms to make it meaningful, but makes it pretty usesless as a job assessment question. A function makes more sense in this context. "There's a variation of the problem which says "find the square root of a double up to the third decimal digit". I'd just multiply the original number by 10^position (meaning, for 3 digit accuracy it'd be 10^3) and ran a simple loop to find the closest integer. As soon as i*i > 10^position * number, you can return (i-1)/10^position. It's hella slow and unimaginative, but it works and you won't sound like you knew this problem ahead of time (if you come up with a Babylonian solution).

Jan 6, 2011

### Software Development Engineer Intern at Amazon was asked...

Jan 25, 2011
 Write a function that takes in an array and repeats an integer that appears the most.5 Answersif: Array [2][2][3][3][3][2][1][2][1] it should print [3]Confusing. In your example, 2 appears the most. Do you mean the integer that repeats the most consecutively? Cause that would be 3. Anyways, in either case, you can go through the array adding all the key-value pairs (number and times) to a hashmap and then access the hashmap in constant time. O(n).class FindMostOccurences { public static DictionaryEntry MostOccurences(int[] Array) { Hashtable ht = new Hashtable(); for (int i = 0; i Int32.Parse(de.Value.ToString())) { { de.Key = item.Key; de.Value = item.Value; } } } return de; } }Show More ResponsesUsing map , i think one loop is sufficient. private static int mostReapeatingNumber(int[] is) { HashMap map = new HashMap(); int tempHighestCount = 0; int keyHighest = 0; for (int index=0; index tempHighestCount) { tempHighestCount = numCount; keyHighest = number; } } } return keyHighest; }I think there's no need to have a map. Just maintain variables prev_max_run, prev_max_num, prev_num, curr_num and curr_run. In the loop if the prev_num was equal to curr_num increment curr_run. When you find the num is different check curr_run with prev_run. If curr_run > prev_run, prev_max_num = curr_num.

Sep 26, 2012