CareerCup

If my understanding is correct, following is the way to solve this.
Taking a simple example of 2 numbers spread over 3 place.
array A = {1,2}; n=3
say the number is: xyz
I see that repetition is allowed, as per the example.
when n=3, we have to take into account n=2 and n=1 based on what I see from the example given in the question.
x can have 1 or 2; y also can have 1 or 2; z also can have 1 or 2.
So the possibilities of all types of 3 digit numbers is: 2*2*2 = 8
Same way, possibilities of all types of 2 digit numbers is 2*2=4
Same way possibilities of all types of 1 digit number is 2
In this case answer would be 14.

Going by this, following snippet would reap the result.

numOfPlaces = 1
while(numOfPlaces<=n) {
result = result + array.length ^ numOfPlaces;
numOfPlaces++;
}

Note: I have not tested the while loop. but conceptually, I believe I have given a fair answer based on the question.
Correct me if wrong.

With a single stack itself this could be done!
Recursion is best.
Enqueue - Usual push operation of a stack
Dequeue - Recursion has to be done, so that the bottom most element is taken from the stack
Listing the elements - Recursion can be done, if you need to list down the elements of queue

Thanks Praveenkumar!
Thanks a lot Ritesh Singh. The link provided by you looks wonderful, though I have just gone through the diagram.

If you are in a service based company, I would not estimate your salary to be that low. May be you could aim for 1-2 lakhs more that what you get now.
But, if you are in a product based company, you are definitely taken for a ride!
Please post here if you get a better opportunity! :-) Thanks!
P.S: This is completely based on my knowledge and interpretations only

So, in Factory pattern, object gets created based on dynamic check... right?
Such dynamic check could also be performed in MVC and objects created. So, what is the uniqueness of factory design pattern?