Engineer Interview Questions

As it is not specified that the International System of Units must be used, define the Elephant unit (E) as the weight of your elephant. Your elephant then weights exactly 1E.

Kyle I have to disagree with you. According to Buoyancy principle: Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. So if you calculate the displaced fluid weight, which is quite easy in water as it is very close or equal to 1kg/L, you can get very close to the real weight of the elephant.

Hmm, I think Kyle is right... and so are you when you say "Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object". IMHO elephants don't float...

Actually Kyle is right. If we put something in the water, such as an sealed box with a ten pound weight inside (does not float), it will displace a proportionate amount of water. However, if we fill it up with a 50 pound weight, the same amount of water will be displaced, however. Fernando, you are using the Buoyancy principle. The Buoyancy principle is meant for objects that float (hence the word buoyancy-it means pressure to keep afloat). As Wouzz said, elephants don't float.

You could just put a boat in water, measure displacement. Then put the elephant in the boat and take the difference. The displacement is going to equal mass as long as it's floating so you're good in a boat. Kyle was right that if you just threw an elephant in a pool, you would have you figure out density and all that.

I would say bring a female elephant and let them enjoy life..

Isn't that just an improvised "weighing machine"?

it sounds like you all are trying to determine whether an elephant is a witch (Monty Python reference). Instead of dealing with the whole mess of a wet elephant, why not use other pneumatic tools like an inflatable platform, an air pump, and a pressure sensor. See how much air pressure is required to lift the elephant.

My experience is this, you don't have to have a perfectly correct answer. The goal of the question is to see if you can think around a problem. Besides elephants, I have heard 747's and aircraft carriers used as the object. A good answer shows that you thought of an alternate way of measurement, so the water displacement and pressure sensor ideas work. PS, if they ask you to weigh a 747, I would answer "Land it on an aircraft carrier and measure the additional water displacement!"

I would use a see saw to weight my elephant. Using my weight (212 lbs in work wear), move the elephant until we balance, and then compare my distance to the pivot point to Clancy's distance to the pivot. (I named my elephant Clancy). The hardest part... finding a see saw strong enough to hold Clancy.

Archimedes Law of the Lever states: Magnitudes are in equilibrium at distances reciprocally proportional to their weights. "Give me a place to stand on, and I will move the Earth."

Elephants DO float, even in fresh water. They're blubbery.

Simple answer : Use a beam balance . Put elephant on one side and start throwing weights on the other side . When the beam is balanced you got the weight of the elephant equal to the sum of weights on the other !

Just ask him...

Just find a person having a bad day and weigh him/her, help that person feel better, then weigh him/her again. The difference is the weight of the elephant you removed off his/her back. Works for monkeys too.

one "weight"(s) by putting a weight on it

note that the answer above is referenced to the question on THIS page which has been transcribed wrongly from the "home page" question (the "t" has been added)

I would ask if it was an Indian elephant or an African elephant.

I would solicit bids and then sub-contract the task to a reputable, cost-effective elephant weight service vendor, and let THEM decide the best approach.They can use a weigh machine or use any other method that does not require ME to use a weigh machine. The question (and answer) are no different than any other variation of, "How do you accomplish a task when you don't have the needed resources?"

by using first clas lever. keep an elephant at known distance from the fulcrum; make the effort distance long enough that your weight balances the weight of an elephant; by using the formula load*load distance = effor*effortdistance( principle of lever) you already know load distance, effort distance and effort now you can calculate load, which is nothing but elephan'ts weight

By using the momentum conservation law, First I find a heavy enough object mith exact mass of m1 and then throw it with speed V1 to the elephant which is standing on a friction-less surface like ice or on plate with wheels under it to reduce the friction. after they meet, the elephant will move. and the object will change its direction and speed. Then I calculate the speed and direction of elephant and reflecting object, and finally I solve the vector equation below: m.v1=m.v2+M.V v1=vector of initial speed of object v2=vector of final speed of object M=elephant mass V=vector of elephant speed

elevate the elephant to height "h". it will have potential energy of "U=mgh". hurl the elephant into a well which is vacumed well to avoid friction. at the end of the well, there is a pool of water with exact mass of " m' " and temprature of T1. after elephant reaches the pool he will give all of his energy to water causing water to become warmer Q=U. so mgh=m'c(T2-T1) with knowing g,h,m',c,T1 and T2 m can be calculated easily.

place elephant on an iron plate and move it horizontally with steady speed v. and pass them through a constant magnetic field of B. change B untill the elephant start to move upward. read B and V. F=V x B is the force of magnetic field ( x denotes vector multiplication) which is now equal to weght force W=mg. knowing g the gravity constant, solve m.

use SEE SAW put elephant on one side and on other side some men when it get balanced weighs the men

Possible answers: 1. Zero. Take the elephant to space. The weight = 0. 2. I can guess it to around 150 kg ( there is fat person who lives next door and we call him an elephant his weight is 150 kg)

I got asked this question. But the water displacement has gotten old. They prompted me for another answer. I said you measure the elephant and find out its volume in cubic meters. Then weigh its leg and measure it. Lets say the leg weighs 20 kgs and has a volume of 20 cubic metres. Then, each cubic metre weighs 1 kg. So if the total volume of elephant is 350 cubic meters, it will weigh 300 cubic metres.

Take heavy weight gym dumbells keep start hanging with the one end of the rope an the other end to elephant and the rope goes over the pully,keep adding weight till elephant gets lifted

esimate the answer to certain range and put it on a boat and if the boat shrinks it is more than the estimate weight else less else use F=MA forumula apply certain force and the calculate acceleration and then mass

OK guys i take a boat which can carry more than a load of an elephant i keep the elephant in that boat [ there fore designed load = load of elephant+ X load] i will add some more load by boarding some swimmers one by one into boat if it starts sinking it means it reached designed load [designed load

By allowing him to sit on us

Fernando,Your answer is really very wierd. First of all: rise in level of water will be equal to elephant's volume not weight. There is a vast difference in volume and weight. Second: if you think that you will estimate it's density to calculate it's weight then it will be a wierd thing because you will never be able to guess it right. Third: rise in level of water of pool will indicate the volume of submerged part of elephant and not whole elephant. If you think that you will submerge whole elephant then obviously you are killing it .So your answer is scientifically and economically wrong. I might suggest that you should take a help of a strong boat to determine elephant weight. take the boat in water and elephant on the boat.the boat will obviously sink a little.mark the level. now get elephant off the boat and put some weights on the boat till it reaches same level. now measure the weights and you will get the weight of elephant THANK YOU

Rohan: What if I give you the string "aaaaaaaaaa"? Your solution will print "a:", not "a10".

in java, if you give me a String str="aaabbbcc"; you can't modify the str as all, then how can you do it in place? I assume I can go through the str, and store the new string such as "a3b3c2" anyone can explain to me?

java inplace public void compress(char[] str) { if (str != null && str.length > 1) { for (int last = 0, curr = 1, count = 1, tail = 0; curr 1) { str[tail++] = str[last]; str[tail++] = (char) count; count = 1; } last = curr; curr++; } } }

Need to delimit the array public void compress(char[] str) { if (str != null && str.length > 1) { int tail = 0; for (int last = 0, curr = 1, count = 1; curr 1) { str[tail++] = str[last]; str[tail++] = (char) count; count = 1; } last = curr; curr++; } str[tail] = '\0'; } }

small correction - public void compress(char[] str) { if (str != null && str.length > 1) { int tail = 0; for (int last = 0, curr = 1, count = 1; curr 1) { str[tail++] = str[last]; str[tail++] = (char) (count + '0'); count = 1; } last = curr; curr++; } str[tail] = '\0'; } }

This solution covers all cases.. http://justprogrammng.blogspot.in/2012/06/amazon-interview-string-compression-c.html

Here is a recursive code which considers all cases... http://justprogrammng.blogspot.com/2012/06/amazon-interview-string-compression-c.html

Try this.. it generates the desired output in Java /** * * @author Sourabh Mishra * Class which handles compression of a string replacing the occurrences of characters * with the character followed by its count in the input string * */ public class StringCompressor { /** * Method takes input as a string and replacing the occurrences of characters * with the character followed by its count in the input string * @param data String * @return */ public String compressString(String data){ StringBuilder outBuilder = new StringBuilder(); char prevChar = data.charAt(0); int counter = 0; char currChar; int length = data.length(); for(int i=0; i< length; i++){ currChar = data.charAt(i); if(currChar == prevChar){ counter++; // For the last unique characters if(i == length-1){ outBuilder.append(currChar); outBuilder.append(counter); } continue; } else { outBuilder.append(prevChar); outBuilder.append(counter); prevChar = currChar; counter=1; } } return outBuilder.toString(); } /** * @param args */ public static void main(String[] args) { StringCompressor sc = new StringCompressor(); System.out.println(sc.compressString("aaabbbcc")); } }

An additional check and write inside in the else class is required to print the final unique character for inputs like "aaabbbh" public class Stringcompress { /** * Method takes input as a string and replacing the occurrences of * characters with the character followed by its count in the input string */ public String compressString(String data) { StringBuilder outBuilder = new StringBuilder(); char prevChar = data.charAt(0); int counter = 0; char currChar; int length = data.length(); for (int i = 0; i < length; i++) { currChar = data.charAt(i); System.out.println("Curr character is :" + currChar); System.out.println("Prev character is :" + prevChar); if (currChar == prevChar) { counter++; // For the last unique characters if (i == length - 1) { outBuilder.append(currChar); outBuilder.append(counter); } continue; } else { System.out.println("Prev character in else is :" + prevChar); outBuilder.append(prevChar); outBuilder.append(counter); prevChar = currChar; counter = 1; if(i==length-1) { outBuilder.append(currChar); outBuilder.append(counter); } } } return outBuilder.toString(); } public static void main(String[] args) { Stringcompress sc = new Stringcompress(); System.out.println(sc.compressString("aaaaaaadfgdfu")); } }

public class Example1 { static String S1="aaabbbccc"; static String comstring=""; public static void main(String[] args) { System.out.println(" Actual string is "+S1); S1=S1+"#"; String oldString=S1; int count=1; for(int i=0;i

import java.io.*; import java.util.*; import java.text.*; import java.math.*; class amzone { public static void main(String [] args) throws Exception { BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); String line=br.readLine(); int n=line.length(); int count=1; for(int i=0;i

Using Ruby methods, ans = "" s.split("").group_by{|x| x}.each{|k,v| ans += "#{k}#{v.count}"} puts ans But this code snippet will produce "a4b3c2" for the input string "aaabbbcca"

import java.util.ArrayList; import java.util.HashMap; import java.util.List; import java.util.Map; public class StringCharCount { public static void main(String args[]) { String inputString="aaabbbccdddddddddddd"; List charUniqueList=new ArrayList(); Map charCountMap=new HashMap(); for(int i=0;i

What about the input "abc"? I assume that should compress to "a1b1c1" which is NOT shorter than the input, so you cannot compress in place.

string CombineWithNoExtraSpace(string S) { int n=S.size(),count=1,i=0; if(i==n) return "\0"; while(i+1<=n && S[i]==S[i+1] ) { count++; i++; } S = S[i] + to_string(count) + combine(S.substr(i+1,n-count)); return S; }

def compress(in_string): final_string = "" init_char = "" ch_val = 1 for (i,ch) in enumerate(in_string): if ch is not init_char: init_char = ch if ch_val > 1: final_string += str(ch_val) ch_val = 1 final_string += ch continue ch_val += 1 try: out = in_string[i+1] except IndexError: final_string += str(ch_val) return final_string print compress("aaaaab")

public String compress(String in) { if (in.length() 1 ? right-left : ""); right++; left = right-1; } return out.toString(); }

public String compress(String in) { if (in.length() 1 ? right-left : ""); right++; left = right-1; } return out.toString(); }

Answer: public class StringCompression { public static void main(String[] args) { String inputString = "abcabcaaaaaaaaaaaaabc";//"aabcccccaaa"; System.out.println(getCompressedString(inputString)); } private static String getCompressedString(String originalString) { StringBuilder compressedStr = new StringBuilder(); if(originalString != null && originalString.length() = originalString.length()){ return originalString; } else{ return compressedStr.toString(); } } }

public static String returnCount (String values) { HashMap getHash = new HashMap(); StringBuilder sb = new StringBuilder(); // char[] charsInString = values.toCharArray(); for(char characters : values.toCharArray()){ if(!(getHash.containsKey(characters))){ getHash.put(characters,1); } else getHash.put(characters,getHash.get(characters) + 1); } // for() for(char keys : getHash.keySet()){ sb.append(keys); sb.append(getHash.get(keys)); } System.out.println(getHash); return sb.toString(); }

#include #include int main(void) { char ch; char accept[100]; char output_string[100]; int count=0,i,j,pos=0; scanf("%s",&accept); printf("%s",accept); for(i=0;i

dfdsf

#include int main() { char c[30]; int i,count=0; printf("Enter a String : "); scanf("%s",c); for(i=0;s[i]!='\0';i++) { count=1; while(s[i]==s[i+1]) { i++; count++; } printf("%s,%d",s[i],count); } return 0; }

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Anonymous, consider the matrix and you are searching for 9. Using this new algorithm we would look at rows 2 and 3 since 8 < 9 < 10 and completely miss 9 in the last row. (1 3 5 7) (2 3 6 8) (3 5 8 10) (6 9 11 12)

dp, what do you do if the first element in the last column is already larger than the target? I.e. there may be no number that is smaller that the target. You could throw away that column, but that's it, I think. If so, the worst case becomes O(m).

What if you did a binary search on the diagonal....

A better solution than O[log(mn)] is unlikely. An mxn array can be laid in memory as a m*n one dimensional array at its base form. So a matrix sorted by row and column simply means this mxn one dimensional array is sorted. A binary search will get you the result in O[log(mn)], that is the fastest solution available.

How about this. for m columns - do binary range_search. As in, start with m/2 th column, if the number is in this range...col(0) O (log(m+n))

oops, there is a catch.... If the number is in a range, it can be in that column, or any of the columns towards left of it....and in some cases, if its in the range.. it can still be on any columns on right..... need to tweak a bit more.

Ok writing this 3rd time , no one will see this but for the sake of it, Using radix sort Take 10 buckets and enter all numbers respectively. For example all numbers unit digit with 9 will in 9 number bucket. Just divide the given number (input) and into parts , using mod operator. And using the divided number"s unit digit find that index in the bucket. Tada . Radix sort way the. Linear search way to find the number Complexity O ( n ) Phew

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Those examples listed are good solutions, but if you wanted to make a bold attempt at wowing them, then here's one answer. This one allows you to save both bulbs and conserve them, which is why I think this answer has a potential wow'ing factor! It's also the answer I would give as a supplemental...it's pretty outlandish for it to stand as your only answer. Using the two bulbs you'll know the make and model. Retrieve the information from the manufacturer that gives data on the physical characteristics, i.e. stress, that the light bulb's glass can take. Then make a calculation that tells you at which height will that stress be reached if you were to drop a bulb from that height. You can also figure out the influence from the material the floor is made of. After that, you go home into your apartment and put those light bulbs back where you found them haha

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.

use they way u use binary search. STEP 1:divide number of floors(n) by 2 = n/2 , drop it there STEP 2: if it breaks, repeat step 1 taking n= n/4 STEP 3: if it doesn not break, drop it next at n=3n/4 refer BINARY SEARCH

@Keshav: What if first bulb breaks at n/2, then according to binary search the next bulb will be dropped at n/4 as u mentioned. What if the second bulb breaks at this point. You dont have any bulbs left to find to out if the threshold is lesser than that????

You start with the n/2 floor. After that you go one by one from either 1 or n/2 +1 until it breaks. If it does not break at n/2 you can go to 3n/4 and try. depending on floor this will either save or possibly add one additional drop. This always gives you Floor + 2 drops (or Floor - n/2 + 2 drops if it is a high floor)

1. Start with the 13th floor. If it breaks, start checking sequentially from floor no. 1 (1, then 2, then 3, and so on) 2. If it doesn't, go to the 13+12th floor = 25th floor. Then to the 25+11th floor = 36. Then 46, 55, and so on. If the egg breaks at, let's say the 55th floor, starting checking for the 46th, 47th, 48th (...) floor till the 54th one. The advantage of using this procedure is you're able to find out the exact floor in a maximum of 13 steps.

I guess the bulb would break from 1st floor itself :p

Assuming no of floors = A, tester can escape x floor at a time e.g. if x = 10, test at 10, 20,30... If 1bulb breaks at 20, test from 11 to 19. So solution is minimum of (A/x + x-1) using derivatives x=√A is for minimum i.e. min value is A/√A +√A -1 =2√A-1. Example for 100 floors minimum is 19

it depends the speed and the force it is doped

Start from floor 1, drop the bulb. If it breaks, then it tells you that at any floor the bulb will break. But, if it doesn't use the same bulb to test its survival on n+1 floor. By continuing this you will reach a floor where the bulb breaks. Hence, you identified the floors where the bulb breaks (BFloor) and the floor where it doesn't (BFloor-1), also you have saved a bulb for use.

Let the size of the string asked is 3 e.g GHD . The formula to this : [ 26+26^2+...+ 26^(size-1) + (LetterIndex-1)*{26^(size-1)+....+(LetterIndex-1)*26}+Index ]

In the code the above formula will use recursion to find the column number.

The answer can be as simple as finding the first combination of number * i such that it is less then xcelcolvalue; ie repeat this till xcelColValue >= 26 ( 0 = 26){ int i = 1, num = 1; while((num * (i + 1)) 26){ i = 1; num++;} else i++; xcelNum -= num*i; sb.append((char) i); } if (xcelNum > 0) sb.append((char) xcelNum); return sb.toString(); }

Also the above code is rough idea and not tested but gives enough idea on simplifying the solution. The code does not use lot of / and % calcuations but could elegantly find the solution. Note the Gaurd if(i > 26){ i = 1; num++;} else i++; which is important in finding the combination of num and i which multiples at least to xcelNum.

I think the above is incorrect and it should be : public static String xcelString(int xcelNum) { StringBuffer sb = new StringBuffer(); while(xcelNum >= 26){ int i = 1; for(;(26 * (i + 1)) 0) sb.append((char) xcelNum); return sb.toString(); }

Edge conditions aside, the program would be /* inputString - the column name. For e.g. "BZC" */ char [] columnName = inputString.toUpperCase().toCharArray(); int columnNumber = 0; for(int i = 0; i < columnName.length - 1; i++){ columnNumber = columnNumber + (int) Math.pow(26, (columnName.length - 1 - i)) * (columnName[i] - 64) ; } columnNumber = columnNumber + (columnName[columnName.length - 1] - 64); System.out.println("Column number is " + columnNumber);

Edge conditions aside, the program would be /* inputString - the column name. For e.g. "BZC" */ char [] columnName = inputString.toUpperCase().toCharArray(); int columnNumber = 0; for(int i = 0; i < columnName.length - 1; i++){ columnNumber = columnNumber + (int) Math.pow(26, (columnName.length - 1 - i)) * (columnName[i] - 64) ; } columnNumber = columnNumber + (columnName[columnName.length - 1] - 64); System.out.println("Column number is " + columnNumber);

int columnNumber(String col) { result = 0; for(i = 0;i< strlen(col) ; i++) { result *= 26; result + = col[i] - 'A' +1; } }

int columnNumber(String col) { result = 0; for(i = 0;i< strlen(col) ; i++) { result *= 26; result + = col[i] - 'A' +1; } }

public static String numToExcel(int n) { if(n 0) { int remainder = n%26; char newChar = (char) ('Z' - (26 -remainder)%26); result = newChar + result; n = (n-1)/26; } return result; }