1. How could Java classes direct program messages to the system console, but error messages, say to a file?
The class System has a variable out that represents the standard output, and the variable err that represents the standard error device. By default, they both point at the system console. This how the standard output could be re-directed:
Stream st =
new Stream (new
FileOutputStream ("techinterviews_com.txt"));
System.setErr(st);
System.setOut(st);
2. What’s the difference between an interface and an abstract class?
An abstract class may contain code in method bodies, which is not allowed in an interface. With abstract classes, you have to inherit your class from it and Java does not allow multiple inheritance. On the other hand, you can implement multiple interfaces in your class.
3. Why would you use a synchronized block vs. synchronized method?
Synchronized blocks place locks for shorter periods than synchronized methods.
4. Explain the usage of the keyword transient?
This keyword indicates that the value of this member variable does not have to be serialized with the object. When the class will be de-serialized, this variable will be initialized with a default value of its data type (i.e. zero for integers).
5. How can you force garbage collection?
You can’t force GC, but could request it by calling System.gc(). JVM does not guarantee that GC will be started immediately.
6. How do you know if an explicit object casting is needed?
If you assign a superclass object to a variable of a subclass’s data type, you need to do explicit casting. For example:
Object a;Customer b; b = (Customer) a;
When you assign a subclass to a variable having a supeclass type, the casting is performed automatically.
7. What’s the difference between the methods sleep() and wait()
The code sleep(1000); puts thread aside for exactly one second. The code wait(1000), causes a wait of up to one second. A thread could stop waiting earlier if it receives the notify() or notifyAll() call. The method wait() is defined in the class Object and the method sleep() is defined in the class Thread.
8. Can you write a Java class that could be used both as an applet as well as an application?
Yes. Add a main() method to the applet.
9. What’s the difference between constructors and other methods?
Constructors must have the same name as the class and can not return a value. They are only called once while regular methods could be called many times.
10. Can you call one constructor from another if a class has multiple constructors
Yes. Use this() syntax.
11. Explain the usage of Java packages.
This is a way to organize files when a project consists of multiple modules. It also helps resolve naming conflicts when different packages have classes with the same names. Packages access level also allows you to protect data from being used by the non-authorized classes.
12. If a class is located in a package, what do you need to change in the OS environment to be able to use it?
You need to add a directory or a jar file that contains the package directories to the CLASSPATH environment variable. Let’s say a class Employee belongs to a package com.xyz.hr; and is located in the file c:/dev/com.xyz.hr.Employee.java. In this case, you’d need to add c:/dev to the variable CLASSPATH. If this class contains the method main(), you could test it from a command prompt window as follows:
c:\>java com.xyz.hr.Employee
13. What’s the difference between J2SDK 1.5 and J2SDK 5.0?
There’s no difference, Sun Microsystems just re-branded this version.
14. What would you use to compare two String variables - the operator == or the method equals()?
I’d use the method equals() to compare the values of the Strings and the = = to check if two variables point at the same instance of a String object.
15. Does it matter in what order catch statements for FileNotFoundException and IOExceptipon are written?
A. Yes, it does. The FileNoFoundException is inherited from the IOException. Exception’s subclasses have to be caught first.
16. Can an inner class declared inside of a method access local variables of this method?
It’s possible if these variables are final.
17. What can go wrong if you replace && with & in the following code:
String a=null;
if (a!=null && a.length()>10)
{...}
A single ampersand here would lead to a NullPointerException.
18. What’s the main difference between a Vector and an ArrayList
Java Vector class is internally synchronized and ArrayList is not.
19. When should the method invokeLater()be used?
This method is used to ensure that Swing components are updated through the event-dispatching thread.
20. How can a subclass call a method or a constructor defined in a superclass?
Use the following syntax: super.myMethod(); To call a constructor of the superclass, just write super(); in the first line of the subclass’s constructor.
21. What’s the difference between a queue and a stack?
Stacks works by last-in-first-out rule (LIFO), while queues use the FIFO rule.
22. You can create an abstract class that contains only abstract methods. On the other hand, you can create an interface that declares the same methods. So can you use abstract classes instead of interfaces?
Sometimes. But your class may be a descendent of another class and in this case the interface is your only option.
23. What comes to mind when you hear about a young generation in Java?
Garbage collection.
24. What comes to mind when someone mentions a shallow copy in Java?
Object cloning.
25. If you’re overriding the method equals() of an object, which other method you might also consider?
hashCode()
26. You are planning to do an indexed search in a list of objects. Which of the two Java collections should you use: ArrayList or LinkedList?
ArrayList
27. How would you make a copy of an entire Java object with its state?
Have this class implement Cloneable interface and call its method clone().
28. How can you minimize the need of garbage collection and make the memory use more effective?
Use object pooling and weak object references.
29. There are two classes: A and B. The class B need to inform a class A when some important event has happened. What Java technique would you use to implement it?
If these classes are threads I’d consider notify() or notifyAll(). For regular classes you can use the Observer interface.
30. What access level do you need to specify in the class declaration to ensure that only classes from the same directory can access it?
You do not need to specify any access level, and Java will use a default package access level.
Friday, October 26, 2007
Friday, October 12, 2007
General Knowledge Questions for Interview
red marbles, blue marbles
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: red marbles, blue marbles
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we'll pick a jar, and each jar will have a 1/2 chance of being picked. then we'll pick a marble, and depending how we stack the marbles, we'll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one.
for example, say we put all the red marbles into jar A and all the blue ones into jar B. then our chances for picking a red one are:
1/2 chance we pick jar A * 50/50 chance we pick a red marble
1/2 chance we pick jar B * 0/50 chance we pick a red marble
do the math and you get 1/2 chance for a red marble from jar A and a 0/2 chance for a red marble from jar B. add 'em up and you get the result = 1/2 chance for picking a red marble.
think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from jar A, but we didn't have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could've just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in jar B to help the odds out there.
let's look at those chances:
1/2 we pick jar A * 1/1 we pick a red marble
1/2 we pick jar B * 49/99 we pick a red marble
do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4.
we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in jar A and maximizing the odds of picking a red marble in jar B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we've maximized the odds in jar A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we've also maximized the number of red marbles in jar B. if we added any more red marbles to jar B we would have to take them out of jar A which reduce the odds there to 0 (very bad). if we took any more blue ones out of jar B we would have to put them in jar A which reduce the odds there by 50% (very bad).
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: red marbles, blue marbles
problem: you have two jars, 50 red marbles, 50 blue marbles. you need to place all the marbles into the jars such that when you blindly pick one marble out of one jar, you maximize the chances that it will be red. (when picking, you'll first randomly pick a jar, and then randomly pick a marble out of that jar) you can arrange the marbles however you like, but each marble must be in a jar.
solution: chance! chance is easy if you know how to do the formula. we know that we have two choices to make. first we'll pick a jar, and each jar will have a 1/2 chance of being picked. then we'll pick a marble, and depending how we stack the marbles, we'll have a (# of red marbles in jar)/(# of total marbles in jar) chance of getting a red one.
for example, say we put all the red marbles into jar A and all the blue ones into jar B. then our chances for picking a red one are:
1/2 chance we pick jar A * 50/50 chance we pick a red marble
1/2 chance we pick jar B * 0/50 chance we pick a red marble
do the math and you get 1/2 chance for a red marble from jar A and a 0/2 chance for a red marble from jar B. add 'em up and you get the result = 1/2 chance for picking a red marble.
think about it for awhile and see if you can figure out the right combination. we had a 50/50 (guaranteed) chance in picking a red marble from jar A, but we didn't have to have 50 red marbles in there to guarantee those fantastic odds, did we? we could've just left 1 red marble in there and the odds are still 1/1. then we can take all those other marbles and throw them in jar B to help the odds out there.
let's look at those chances:
1/2 we pick jar A * 1/1 we pick a red marble
1/2 we pick jar B * 49/99 we pick a red marble
do the math and add them up to get 1/2 + 49/198 = 148/198, which is almost 3/4.
we can prove these are the best odds in a somewhat non-formal way as follows. our goal is to maximize the odds of picking a red marble. therefore we can subdivide this goal into maximizing the odds of picking a red marble in jar A and maximizing the odds of picking a red marble in jar B. if we do that, then we will have achieved our goal. it is true that by placing more red marbles into a jar we will increase the chances of picking a red marble. it is also true that by reducing the number of blue marbles in a jar we will increase the odds also. we've maximized the odds in jar A since 1/1 is the maximum odds by reducing the number of blue marbles to 0 (the minimum). we've also maximized the number of red marbles in jar B. if we added any more red marbles to jar B we would have to take them out of jar A which reduce the odds there to 0 (very bad). if we took any more blue ones out of jar B we would have to put them in jar A which reduce the odds there by 50% (very bad).
General Knowledge Questions for Interview
100 doors in a row
100 doors in a row aha:!
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
question: what state are the doors in after the last pass? which are open which are closed?
solution: 100 doors in a row
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
for example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8...) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..
question: what state are the doors in after the last pass? which are open which are closed?
solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9... leaving it open at the end. only perfect square doors will be open at the end.
100 doors in a row aha:!
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
question: what state are the doors in after the last pass? which are open which are closed?
solution: 100 doors in a row
problem: you have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.
for example, after the first pass every door is open. on the second pass you only visit the even doors (2,4,6,8...) so now the even doors are closed and the odd ones are opened. the third time through you will close door 3 (opened from the first pass), open door 6 (closed from the second pass), etc..
question: what state are the doors in after the last pass? which are open which are closed?
solution: you can figure out that for any given door, say door #42, you will visit it for every divisor it has. so 42 has 1 & 42, 2 & 21, 3 & 14, 6 & 7. so on pass 1 i will open the door, pass 2 i will close it, pass 3 open, pass 6 close, pass 7 open, pass 14 close, pass 21 open, pass 42 close. for every pair of divisors the door will just end up back in its initial state. so you might think that every door will end up closed? well what about door #9. 9 has the divisors 1 & 9, 3 & 3. but 3 is repeated because 9 is a perfect square, so you will only visit door #9, on pass 1, 3, and 9... leaving it open at the end. only perfect square doors will be open at the end.
General Knowledge Questions for Interview
reverse a string - word by word
a typical programming interview question is "reverse a string, in place". if you understand pointers, the solution is simple. even if you don't, it can be accomplished using array indices. i usually ask candidates this question first, so they get the algorithm in their head. then i play dirty by asking them to reverse the string word by word, in place. for example if our string is "the house is blue", the return value would be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: reverse a string - word by word
problem: reverse "the house is blue", the answer should be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: solving the initial problem of just reversing a string can either be a huge help or a frustrating hinderance. most likely the first attempt will be to solve it the same way, by swapping letters at the front of the string with letters at the back, and then adding some logic to keep the words in order. this attempt will lead to confusion pretty quickly.
for example, if we start by figuring out that "the" is 3 letters long and then try to put the "t" from "the" where the "l" from "blue" is, we encounter a problem. where do we put the "l" from "blue"? hmm... well we could have also figured out how long "blue" was and that would tell us where to put the "l" at... but the "e" from "blue" needs to go into the space after "the". argh. its getting quite confusing. in fact, i would be delighted to even see a solution to this problem using this attack method. i don't think its impossible, but i think it is so complex that it's not worth pursuing.
here's a hint. remember before when we just reversed "the house is blue"? what happened?
initial: the house is blue
reverse: eulb si esuoh eht
look at the result for a minute. notice anything? if you still don't see it, try this.
initial: the house is blue
reverse: eulb si esuoh eht
wanted : blue is house the
the solution can be attained by first reversing the string normally, and then just reversing each word.
a typical programming interview question is "reverse a string, in place". if you understand pointers, the solution is simple. even if you don't, it can be accomplished using array indices. i usually ask candidates this question first, so they get the algorithm in their head. then i play dirty by asking them to reverse the string word by word, in place. for example if our string is "the house is blue", the return value would be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: reverse a string - word by word
problem: reverse "the house is blue", the answer should be "blue is house the". the words are reversed, but the letters are still in order (within the word).
solution: solving the initial problem of just reversing a string can either be a huge help or a frustrating hinderance. most likely the first attempt will be to solve it the same way, by swapping letters at the front of the string with letters at the back, and then adding some logic to keep the words in order. this attempt will lead to confusion pretty quickly.
for example, if we start by figuring out that "the" is 3 letters long and then try to put the "t" from "the" where the "l" from "blue" is, we encounter a problem. where do we put the "l" from "blue"? hmm... well we could have also figured out how long "blue" was and that would tell us where to put the "l" at... but the "e" from "blue" needs to go into the space after "the". argh. its getting quite confusing. in fact, i would be delighted to even see a solution to this problem using this attack method. i don't think its impossible, but i think it is so complex that it's not worth pursuing.
here's a hint. remember before when we just reversed "the house is blue"? what happened?
initial: the house is blue
reverse: eulb si esuoh eht
look at the result for a minute. notice anything? if you still don't see it, try this.
initial: the house is blue
reverse: eulb si esuoh eht
wanted : blue is house the
the solution can be attained by first reversing the string normally, and then just reversing each word.
General Knowledge Questions for Interview
salary
three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries?
solution: salary
How about: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.
As has been noted already, there's no way to liar-proof the scheme.
It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.
three coworkers would like to know their average salary. how can they do it, without disclosing their own salaries?
solution: salary
How about: Person A writes a number that is her salary plus a random amount (AS + AR) and hands it to B, without showing C. B then adds his salary plus a random amount (BS + BR) and passes to C (at each step, they write on a new paper and don't show the 3rd person). C adds CS + CR and passes to A. Now A subtracts her random number (AR), passes to B. B and C each subtract their random number and pass. After C is done, he shows the result and they divide by 3.
As has been noted already, there's no way to liar-proof the scheme.
It's also worth noting that once they know the average, any of the three knows the sum of the other 2 salaries.
General Knowledge Questions for Interview
world series
you have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).
unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?
basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.
solution: world series
this probably isn't the cleanest solution, but...
a dynamic-programming type solution is:
(1) Create a 5x5 matrix P.
So, P[i,j] holds your pile of money when the yanks have won i games and the mets have won j games.
initialize P[4,j] := 20 for j from 0 to 3 initialize P[i,4] := 0 for i from 0 to 3
fill P in bottom-right to top left by averaging bottom and right adjacent cells:
P[i,j] := (P[i+1,j]+P[i,j+1]) / 2
(2) Make another 5x5 matrix, B, which represets your bet at any-time.
So, B[i,j] represents your bet when the yanks have won i games and the Mets j games.
fill this top-left to bottom right by:
B[i,j] = P[i+1,j] - P[i,j]
(3) Look in matrix B for your bet at any time.
The final matricies are:
Pile-Matrix
0.00
1.00
2.00
3.00
4.00
0
10.00
6.88
3.75
1.25
0.00
1
13.13
10.00
6.25
2.50
0.00
2
16.25
13.75
10.00
5.00
0.00
3
18.75
17.50
15.00
10.00
0.00
4
20.00
20.00
20.00
20.00
Bet-Matrix
0
1
2
3
4
0
3.13
3.13
2.50
1.25
1
3.13
3.75
3.75
2.50
2
2.50
3.75
5.00
5.00
3
1.25
2.50
5.00
10.00
4
you have $10,000 dollars to place a double-or-nothing bet on the Yankees in the World Series (max 7 games, series is over once a team wins 4 games).
unfortunately, you can only bet on each individual game, not the series as a whole. how much should you bet on each game, so that, if the yanks win the whole series, you expect to get 20k, and if they lose, you expect 0?
basically, you know that there may be between 4 and 7 games, and you need to decide on a strategy so that whenever the series is over, your final outcome is the same as an overall double-or-nothing bet on the series.
solution: world series
this probably isn't the cleanest solution, but...
a dynamic-programming type solution is:
(1) Create a 5x5 matrix P.
So, P[i,j] holds your pile of money when the yanks have won i games and the mets have won j games.
initialize P[4,j] := 20 for j from 0 to 3 initialize P[i,4] := 0 for i from 0 to 3
fill P in bottom-right to top left by averaging bottom and right adjacent cells:
P[i,j] := (P[i+1,j]+P[i,j+1]) / 2
(2) Make another 5x5 matrix, B, which represets your bet at any-time.
So, B[i,j] represents your bet when the yanks have won i games and the Mets j games.
fill this top-left to bottom right by:
B[i,j] = P[i+1,j] - P[i,j]
(3) Look in matrix B for your bet at any time.
The final matricies are:
Pile-Matrix
0.00
1.00
2.00
3.00
4.00
0
10.00
6.88
3.75
1.25
0.00
1
13.13
10.00
6.25
2.50
0.00
2
16.25
13.75
10.00
5.00
0.00
3
18.75
17.50
15.00
10.00
0.00
4
20.00
20.00
20.00
20.00
Bet-Matrix
0
1
2
3
4
0
3.13
3.13
2.50
1.25
1
3.13
3.75
3.75
2.50
2
2.50
3.75
5.00
5.00
3
1.25
2.50
5.00
10.00
4
General Knowledge Questions for Interview
100 factorial aha:!
how many trailing zeroes are there in 100! (100 factorial)?
solution: 100 factorial
One per factor of 10, and one per factor of 5 (there are more than enough 2's to pair with the 5's), plus one per factor of ten squared (one occurrence) and one per factor of 5 squared (three occurrences).
So if I'm counting correctly, that'd be 10 + 10 + 1 + 3== 24 zeroes.
Assuming the question meant *trailing* zeroes. It'd be much harder to also count the intermingled zero digits in the entire expansion.
how many trailing zeroes are there in 100! (100 factorial)?
solution: 100 factorial
One per factor of 10, and one per factor of 5 (there are more than enough 2's to pair with the 5's), plus one per factor of ten squared (one occurrence) and one per factor of 5 squared (three occurrences).
So if I'm counting correctly, that'd be 10 + 10 + 1 + 3== 24 zeroes.
Assuming the question meant *trailing* zeroes. It'd be much harder to also count the intermingled zero digits in the entire expansion.
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