Graduate Interview Questions

Assumptions: You have three inputs: X, Y, Z with two possible states (0,1). You have three outputs: A,B,C with two possible states (0,1). The inputs and outputs are independent of each other, but one input should be matched to one output. For example, if X --> A, then X(1) --> A(1). Note: we don't actually know whether the normal state of the switch is open or closed. For our purposes 0 is normal state, 1 is abnormal state. You have one chance to observe the outputs. You could flip one of the switches and observe, but this will only tell you about one of the connections and leave two unknowns. You could flip two switches, but this still leaves two unknowns. You could flip all three, but this leaves three unknowns. You could flip none of them, but that leaves three unknowns. Another possibility is to McGyver the light switches by placing three resistors of different values in series with the switch, verifying the switch is closed, and then observe the relative brightnesses of the bulbs. More creatively, you could use your one trip to leave the room to find the documentation and/or the persons responsible for this cruel experiment. I am sure this is wrong, but I was never a math or CS major. =D

(I mean you would put one resistor in series with each of the three switches, obviously; the greatest value resistor would result in the most dimming).

My answer to this is: Assign a value to each light switch (A,B,C) Start a timer for 3 minutes Flip switch A and B on After the 3 minutes is up, flip B off again. Go into the other room. The lighting bulb belongs to switch A The the other two light bulbs (one will be hot and one will be cold) The hot bulb belongs to switch B The cold bulb belongs to switch C.

I passed my exam in 2006 & I have B Tech ( Hons ) Degree

I got my DAE Diploma ( Civil ) In 1996 at First Division

This was more of a pair programming styled interview. Very nice and collaborative.

public class Words { public static String[] dic = new String[80368]; public static void split(String head, String in) { // head + " " + in is a segmentation String segment = head + " " + in; // count number of dictionary words int count = 0; Scanner phraseScan = new Scanner(segment); while (phraseScan.hasNext()) { String word = phraseScan.next(); for (int i=0; i

There is no efficient solution, just focus on solving the problem first. Use a hashtable to keep track of values above and to the left, then have a while loop that finds the lowest value.

Assume it's a 3x3 matrix, start filling the matrix from 9 to 1 from left to right. |9 8 7| |6 5 4| |3 2 1| If it was a 4x4 array, start from 16 to 1.

I would call a coffee break and address each of the colleagues separately to try to determine the root of the problem. I would then set up a resolution discussion in a bid to resolve it.

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Switch on switches 1 & 2, wait a minute and switch off number 2. Enter the room. Whichever bulb is on is wired to switch 1, whichever is off and hot is wired to switch number 2, and the third is wired to switch 3.

turn on two switches for few minutes and leave one off. Then, right before you leave turn off another one so that one switch is up and the other 2 are down. Obviously the on light bulb corresponds to the high switch. The off but still warm light bulb corresponds to the switch that was just turned off. The other off bulb corresponds to the other switch

Implement a way of iterating through an array of integers and returning only duplicate values within the array, wasn't able to answer this one but did give me a hint abut using hashmaps to do it

import java.util.*; public class GetDuplicates { public static void main(String args[]){ // get input Scanner sc = new Scanner(System.in); ArrayList nums = new ArrayList(); while(sc.hasNext()) { String num = sc.next().trim(); if (num.equals("q")) break; nums.add(Integer.parseInt(num)); } Set set = new HashSet(); for(int n : nums) { if(set.contains(n)) { System.out.print(n + " "); } else { set.add(n); } } } }