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

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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

Applications Developer was asked...May 28, 2015

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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

Applications Developer was asked...February 26, 2016

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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

Applications Developer was asked...September 11, 2011

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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

Applications Developer was asked...July 11, 2012

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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

Senior Applications Developer was asked...February 14, 2011

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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

An Applications for the Post of Software Engineer was asked...October 15, 2013

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When ever I get a work I never leave that work in between before completing it. This is one of my weakness. Less

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i lose my temper quickly

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go to their official sites..

Applications Developer was asked...August 21, 2016

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A 3x3 2D array?, just store the sequence in which the blocks have to be selected

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Graph data structure can be used, which consists of vertices, edges and x,y relation between vertices Less

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graph (gesture) datastructure

Applications Developer was asked...March 23, 2016

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var MoveZerosToRight = function(arr) { let n = arr.length; let i = 0; for(let num of arr) { if(num !== 0){ arr[i] = num; i++; } } while(i < n ) { arr[i] = 0; i++; } return arr; }; Less

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Used 2 pointers one for zeroes and other for numbers, occasionally swapping to achieve what I want. Less

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How will you be achieving the same order?

Senior Applications Developer was asked...February 14, 2011

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Weigh these together: 10 coins from stack 1, 9 from stack 2, etc ending with 1 from stack 10. The weight of these will tell you which stack has the 2g coins. Ex: if it's 1st stack: 65g, 2nd: 64g, 10th: 54g Less

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A more eloquent answer would be: Weigh together 1 coin from stack 1, 2 coins from stack2, 3 coins from stack 3, etc. Subtract 55 from that total weight to get the number of the stack with the 2g coins. Less

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Excellent question!