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Applications engineer Interview Questions in United States

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IBM
Applications Developer - Business Intelligence was asked...January 23, 2011

What reporting tools have you used? Actuate, Brio, Crystal Reports? How much experience have you had in scheduling jobs on Unix? Can you write XML and implement it?

1 Answers

The interviewer was interested in breath of experience rather than specific technical skill. Unix work with scheduling and XML skills are necessary, but not heavy (the same procedures are done over and over again). Emphasis was on selling yourself through experience. Less

Bank of America

I was asked about XSS, SQL Injection, Tools I have used for pen testing.

1 Answers

Cross-site scripting (XSS) is a type of computer security vulnerability typically found in Web applications. XSS enables attackers to inject client-side script into Web pages viewed by other users. A cross-site scripting vulnerability may be used by attackers to bypass access controls such as the same origin policy. - SQL injection is a code injection technique, used to attack data-driven applications, in which malicious SQL statements are inserted into an entry field for execution (e.g. to dump the database contents to the attacker). Less

Altera

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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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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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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Deutsche Bank

We have a pond containing a single bacterium. The number of bacteria double every 5 minutes, and the pond is full of them in 24 hours. If we started with the same pond but two bacteria, how long will it take to fill the pond?

4 Answers

I struggled with this a bit and got close. I believe answer is: 23:55

The first pond started with 1 bacterium and doubled to 2 in five minutes. Therefore, the second pond will take 5 minutes less than the first to be full. ie: 23:55 Less

This is a clear case of Geometric progression. Find the nth term Tn1 = a*r^(n-1). where n = (24 * 60)/5,a = 1 and r=2. when the initial value (a) = 2, the values become n = ?, a = 2 and r = 2. Since Tn1 = Tn2, Equate the RHS of both the equation. Since the base are equal, equate the powers, doing so will give the n value. When n is convert into minutes one get 23 hrs 55 minutes. Less

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