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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
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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
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Nop Not yet.. Still waiting.. Good luck for you alex.. Let me know if in case you get any response.. Less
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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
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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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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
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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
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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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I wont give the answer. Hint : know who divides you ;-)
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I think 2520 is perfect
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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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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
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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
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Thanks man, best wishes to you too.. Nail it this time
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i got it to be in the 14 th century ...
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Sorry it should be 08-31-1380
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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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How was the phone interview? you didn't have on-site interview?
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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
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can we choose any two language out of 5.?
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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
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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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import java.util.HashMap; import java.util.Map; import java.util.Scanner; public class Test { public static void main(String[] args){ Map map = new HashMap(); Scanner s1 = new Scanner(System.in); int a[] = new int[10]; System.out.println("Please enter the 10 numbers"); for(int i=0;i<10;i++){ a[i]=s1.nextInt(); } for(int j=0;j<10;j++){ int count = 1; for(int k=0;k<10;k++){ if(j!=k){ if(a[j]==a[k]){ count++; } } } map.put(a[j],count); } for(Map.Entry m:map.entrySet()){ System.out.println("The value "+m.getKey()+" occurs "+m.getValue()+" times in the series which u have entered."); } } } Less
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I struggled with this a bit and got close. I believe answer is: 23:55
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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
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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