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Applications support analyst Interview Questions

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Altera
Applications Engineer was asked...April 30, 2015

Interview was with a Senior Application Engineer-Hiring manager. The process was: --Self Introduction, --Work experiences, --Go over your resume and your educational experiences/projects in digital system --basic questions on Synthesis flow

37 Answers

Nop Not yet.. Still waiting.. Good luck for you alex.. Let me know if in case you get any response.. Less

I had applied for the position which was posted for munich but during the Interview I was told that the training will be in the UK irrespective of the location you apply for. Less

Ok Guys, I have got a feedback from Altera. They have invited me for a personal Interview to the UK. I think Munich was not feasible for them. I hope you all have got a response too.. Less

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Oracle

You have 2 light bulbs. You are in multistory building such that if you drop a bulb from a floor of a certain height or higher the bulb will break: for ex: if the bulb will break at a minimum height of 10th floor, then the bulb will break if dropped only if dropped from a floor higher than tenth floor. it will not break if dropped from ninth floor or less. using the two bulbs how will you figure out at which floor will the bulb break?

14 Answers

I forgot to mention that you have to try to figure out the floor in the least amount of tries. so incase the minimum floor is 100, you shouldn't try all the floors. The solution posted above works, but is the least efficient. Less

With 2 bulbs we can skip 3 floors at a time. 1) Drop bulb #1 at floor x 2) If it breaks then try floor x-1 else try floor x+3 Start with floor 1. Less

The issue with a research-based answer is that parameters of the posed problem is not based in reality, i.e. dropping a light bulb from any floor would almost guarantee it breaking in real life. So the interviewer is really looking for problem solving skills for an hypothetical situation rather your attention to detail in a real world situation. Less

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Thoughtworks

10 people can share a bucket of coins equally. A monkey steals one coin. The no of coins are one less than equal share. one person after the other tries to take the coin but monkey kills them(killing spree?? :-)). each time a person dies the no of coins are always one short of equal share. what were the no of coins originally?.

11 Answers

I wont give the answer. Hint : know who divides you ;-)

I think 2520 is perfect

Answer is pretty simple: m%10=0 -----case 1 Lets assume n=m-1 Since the remainder is always same , when n is divider by any number from 10 to 1.. n%10 = n%9=.........=1 That means n+1 is completely divisible by all numbers from 10 to 1.. (Since it gives common remainder).. n + 1= Lcm of (1 to 10) = 2520 But we assumed n = m-1, so m -1+1 =m = 2520 Less

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J.P. Morgan

1. Take an integer input and output the number of 1's in it's binary representation. 2. Implement a mergesort. 3. Explain your level of understanding of data structures (trees, etc.) 4. What makes java different than other languages?

9 Answers

Hey, did you hear back from them about the final decision? Also, were u coding in Java or Python? If python, why did they ask you questions on java? Could u share those questions because I have been invited too Less

They ask you what language you feel most comfortable with and ask you questions based on that (I said Java). I haven't heard back but I have been contacted by another recruiter from their Manhattan office so maybe better luck on this second try. Good Luck! Less

Thanks man, best wishes to you too.. Nail it this time

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Active Network

Most unexpected question: Describe to me the process and benefits of wearing a seatbelt.

9 Answers

Process: put buckle into socket until it locks. Benefit: your face doesn't become shredded to hell when it goes flying through the windshield in an accident. Less

Process (front seat manual seat belt only for simplicity): Prerequisites: - Be inside car. - Be seated. - Be unbuckled. Step 1: move the hand farthest from your nearest window over your opposite shoulder, reaching just behind the window frame for the belt buckle. Step 2: Grab it and pull your hand across your body in a downward fashion, reaching for the space between your seat and the seat next to you, where you'll find the socket into which the buckle locks. Step 3: Match the hole in the socket with the buckle. It may take some practice. Step 4: CLICK! Benefits (from most to least likely to happen): - Avoiding minor injuries from hitting your head against the windshield when the person driving brakes too hard all of a sudden. - Being able to slide your nail against the side of the belt, making a "swish" noise which will potentially annoy everyone else inside the car, for endless fun. - Achieving a false sense of security. - Avoiding fines for not using it (where applicable). And not having to use one of these: http://static2.stuff.co.nz/1368495490/264/8670264.jpg - If you do end up having a real car crash, you won't have to worry about any nasty sequels, as you'll likely die caught in its relentless grip as your car explodes. - Avoiding dying when the car comes to an abrupt stop after having gathered enough momentum to throw you off through the windshield. Less

flight attendant, assume crash position...

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Oracle

Tell me a palindrome date before 10-02-2001 (mm-dd-yyyy)

7 Answers

i got it to be in the 14 th century ...

Sorry it should be 08-31-1380

how can the month no. will be more than 12 i think the answer should be 29-11-1192 :) correct me if i am wrong..... Less

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MathWorks

2nd Round (1 hour) Programming related question on java and c/c++ Each section has 8 questions (2 programming and 6 multiple choice) 3rd Round - Phone Interview

7 Answers

How was the phone interview? you didn't have on-site interview?

Phone interview was good. They asked the questions which are there on the blackboard and there a set of mathwork interview questions on scribd.com. You can also refer that Less

can we choose any two language out of 5.?

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MathWorks

What is a null pointer?

6 Answers

I guess what the interview want to hear is that "do not forget to release the memory the pointer pointed to before set it to null" Less

Once the memory is release using the delete command, the pointer is pointing to null. Now if we try to delete it again, the heap will be corrupt. Less

Definition: A null pointer is a general pointer (int *i, char *c, etc..,) except that the address stored in i, c doesn't correspond to any memory location. (i=0, c=0) Less

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How fast u can join the company to work

6 Answers

Immediately

Asap. I am on my notice period. So I can join immediately

I'm eagerly waiting to do this job

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Zoho

Write a Java program to count the number of occurances of each number in a series of numbers.

5 Answers

public static void main(String[] args) { int occurrence = 0; Map m = new HashMap(); List nos = new LinkedList(); nos.add(1); nos.add(1); nos.add(2); nos.add(3); nos.add(4); nos.add(3); nos.add(2); nos.add(4); nos.add(5); nos.add(100); Set uniqueNos = new HashSet(nos); for(int temp:uniqueNos) { for(int tempList:nos) { if(temp == tempList) { occurrence++; } } m.put(temp, "Appears "+occurrence +" times"); occurrence = 0; } System.out.println(m); } Less

public class FirstJava { public static void main(String[] args) { int occurrence = 0; List nos = new LinkedList(); nos.add(1); nos.add(1); nos.add(2); nos.add(3); nos.add(4); nos.add(3); nos.add(2); nos.add(4); nos.add(5); nos.add(100); Map uniqueNumberCount = new HashMap(); for(int datano:nos){ int k=1; if(uniqueNumberCount.containsKey(datano)){ k=uniqueNumberCount.get(datano); } uniqueNumberCount.put(datano, k); } for(Map.Entry unique:uniqueNumberCount.entrySet()){ System.out.println(unique.getKey()+" Unique Value --->"+unique.getValue()); } } } Less

import java.util.Scanner; public class Occurrence { public static void main(String[] args) { int n, count = 0, i = 0; Scanner s = new Scanner(System.in); System.out.print("Enter no. of elements you want in array:"); n = s.nextInt(); int a[] = new int[n]; int x[] = new int[n]; System.out.println("Enter all the elements:"); for(i = 0; i < n; i++) { a[i] = s.nextInt(); } System.out.print("Enter the element of which you want to count number of occurrences:"); for(i = 0; i < n; i++) { x[i] = a[i]; } for(i = 0; i < n; i++){ count = 0; for(int j = 0; j < n; j++) { if(a[j] == x[i]) { count++; } } System.out.println("Number of Occurrence of the Element:"+a[i]+"is"+count); } } } Less

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