Solution: 

Bottom-up parsing is a strategy for analyzing unknown data relationships that attempts to identify the most fundamental units first, and then to infer higher-order structures from them. It attempts to build trees upward toward the start symbol. It occurs in the analysis of both natural languages and computer languages. 

Top-down parsing is a strategy of analyzing unknown data relationships by hypothesizing general parse tree structures and then considering whether the known fundamental structures are compatible with the hypothesis. It occurs in the analysis of both natural languages and computer languages. Please refer to these links for much better information. 

http://en.wikipedia.org/wiki/Bottom-up_parsing 

http://en.wikipedia.org/wiki/Top-down_parsing

Solution: 
In computer science, a symbol table is a data structure used by a language translator such as a compiler or interpreter, where each identifier in a program’s source code is associated with information relating to its declaration or appearance in the source, such as its type, scope level and sometimes its location.
Check out 
http://en.wikipedia.org/wiki/Symbol_table

Solution:
Every row has a primary key. Suppose the primary key for this
particular database is the name of the user then we can sort the names based
on alphabets and do secondary indexing based on the starting alphabet . If
the data is uniformly distributed we can go for multilevel indexing or
hashing.Similarly if we have a registration number as the primary key then
we can sort the table based on registration number and then do indexing
either secondary level or multilevel or apply hashing techniques based on
the distribution of data. Many efficient algorithms are available for
indexing and hashing.

Solution:
Weigh 3 balls against 3 others. 
Case A: If, on the first weighing, the balls balance, then the defective is among the 2 remaining balls and can be determined using 2 weighings making it a total of 3.

Case B:

Step1: If, on the first weighing, the balls don’t balance.
If the balls do not balance on the first weighing, we know that the odd ball is one of the 6 balls that was weighed. We also know that the group of 2 unweighed balls are normal, and that one of the sides, let’s say Side A, is heavier than the other (although we don’t know whether the odd ball is heavy or light). 
Step 2 : Take 2 balls from the unweighed group and use them to replace 2 balls on Side A (the heavy side). Take the 2 balls from Side A and use them to replace 2 balls on Side B (which are removed from the scale). 

I. If the scale balances, we know that one of the 2 balls removed from the scale was the odd one. In this case, we know that the ball is also light. We can proceed with the third weighing amd determine the lighter of the 2 balls ,hance the defective.

II. If the scale tilts to the other side, so that Side B is now the heavy side, we know that one of the three balls moved from Side A to Side B is the odd ball, and that it is heavy. We proceed with the third weighing and determine the heavier one ,the defective. 

III. If the scale remains the same, we know that one of the two balls on the scale that was not shifted in our second weighing is the odd ball. We also know that the unmoved ball from Side A is heavier than the unmoved ball on Side B (though we don’t know whether the odd ball is heavy or light).

Step 3 (for Case B): Weigh the ball from Side A against a normal ball. If the scale balances, the ball from Side B is the odd one, and is light. If the scale does not balance, the ball from Side A is the odd one, and is heavy.

Solution:
Dozens of different data structures have been proposed for implementing dictionaries including hash tables, skip lists, and balanced/unbalanced binary search trees — so choosing the right one can be tricky. Depending on the application, it is also a decision that can significantly impact performance. In practice, it is more important to avoid using a bad data structure than to identify the single best option available.As the frequency of look ups for a word is also important,weighted binary search tree with weights in proportion to the frequency of lookups and determining the depth, can be effective.

Solution:

A zero in n! typically occurs when a multiple of 5 gets multiplied to an even number.We use this simple yet effective information to solve this problem.In the first n natural numbers,those divisible by 5 are always less than the no of even numbers.So it all boils down to the power of 5 in the prime factorization of n! .
This simple formula works for finding it floor(n/5)+floor(n/25)+floor(n/125)+……

function zeros(int n)
{
  int count=0,k=5;
while(n>=k)
{
 count+=n/k;
        k*=5;
}
return count;
}

this count is the number of o’s in n!.