1. What is Bottom up parsing and what is top down parsing ?

Answer :  

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.

2. How many times a day does a clock’s hands overlap?

Answer :

The Hour hand and Minute hand would be meeting exactly 11 times in 12 hours (Hour hand would have taken 1 clockwise round and Minute hand would have taken 12 clockwise rounds, so 12 - 1 = 11 rounds).

result: First time hour and minute hands overlap will be 12 Hours / 11 = 01:05:27.27. So at this time only hour and minute hands would be overlapping and second hand will not be any near to them. Similarly for 2nd, 3rd, 4th, 5th, 6th, 7th, 8th, 9th and 10th overlap of hour and minute hand the Second hand wont be any nearby. So all 3 hands (hour, minute and Second) overlap only 2 times i.e. (0:0:0 and 12:0:0).

Also we all know when we get our watches repaired, normally the repairman overlaps all the three hands to 12.

If we are considering that the second hand is not present, then the rest two overlaps 22 times in 24 hours.

There again is a catch, if we check the angles by which the hour hand and minute hand moves.

The second hand moves 6 degree in a second. In that time the minute hand will move 6/60 degrees. and the hour hand will move 6/(60*12) degrees. now taking these things in the considerations. if we check the positions of the hour and minute hand in terms of angle from the marker 12, for our first rendezvous time, i.e. 01:05:27.27 sec.
first thing that comes to my mind is that, there is fraction in the seconds. So that time can’t be measured. there will be no exact overlap. now lets calculate the angles:

1 hour 5 mins and 27 seconds = 3600 + 5*60 + 27 = 3927 seconds.

angle of hour hand = 3927 * 6/(60*12) = 32.725 degree.
angle of minute hand = 3927 * 6/60 = 392.7 degree
subtracting 360 degree from it we get - 32.7 degree.

So at 01:05:27 both hands don’t overlap. Now for 01:05:28 :
Angles : hour hand - 32.73333
minute hand - 32.8
so obviously they dont meet at 01:05:28 either.

So they overlap at 12:00 and 24:00 only. So the answer is 2 only.

3. If array A is small and array B is too large. how will you proceed for getting intersection of those two arrays?

Answer :   In this case for each entry of smaller array,we can run a binary search routine on the larger one to know its presence.

4. If the probability of observing a car in 30 minutes on a highway is 0.95, what is the probability of observing a car in 10 minutes (assuming constant default probability)?

Answer :

If the chance to see the car is 10 percent per minute, the first minute you have 10% chance, the second minute you have 10% of 90% = 9% (so total 19%), the third minute 10% of 81% (= 8,1%, total 27,1 %) ......
As the chance for 30 minutes is 95 percent, the chance for 1 minute is 9.5% and for 10 minute 63.1 %.

5. What is a symbol table?

Answer :

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.

6. What are dangling pointers?

Answer :

A dangling pointer is a pointer to storage that is no longer allocated. Dangling pointers are nasty bugs because they seldom crash the program until long after they have been created, which makes them hard to find. Programs that create dangling pointers often appear to work on small inputs, but are likely to fail on large or complex inputs.

7. Write code for finding number of zeros in n!

Answer :

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=1;while(n>=pow(5,k)){ count+=n/pow(5,k);        k++;}return count;}
this count is the number of o's in n!.

8. You have five pirates, ranked from 5 to 1 in descending order. The top pirate has the right to propose how 100 gold coins should be divided among them. But the others get to vote on his plan, and if fewer than half agree with him, he gets killed. How should he allocate the gold in order to maximize his share but live to enjoy it? (Hint: One pirate ends up with 98 percent of the gold.)

Answer :

The highest ranked pirate gets 98 gold coins
---Two pirates get 1 gold coin each
---The other 2 pirates get nothing.

9. You have cycle in linked list. Find it. Prove that time complexity is linear. Also find the node at which looping takes place.

Answer :

The problem of checking whether there is a cycle or not can be solved using 2 pointers one moving in increments of 1 and the other in increments of 2.If there is a cycle then these 2 pointers meet at some node say N1 inside the cycle otherwise the fast pointer reaches the end of the list.This is a O(N) solution.

Now coming to the identification of the node at which looping took place.After our identification of cycle ,both the pointers P1 and P2 are at node N1.Now iterate the slow pointer to count the no of nodes in the cycle.(After traversing the whole cycle P1 and P2 shall again be at the same node).Let this size be K.Now take one of the pointers to the head node and count the no of nodes till N1.Let this number be X.Now use one of these pointers to reverse the cycle starting from N1.Only the cycle gets reversed.Now again traverse from head node to N1.Let the number of nodes this time be Y.Let the no of nodes from head to the start node of the cycle be Z

Now X+Y=2*Z+K .Hence solve for K and then having figured out the start node N2 of the cycle.Now as the cycle is reversed having figured out this start node its next node is the looping nodes so set the looping nodes next pointer to NULL and reverse the list further till you reach N2.

10. Four people need to cross a rickety rope bridge to get back to their camp at night. Unfortunately, they only have one flashlight and it only has enough light left for seventeen minutes. The bridge is too dangerous to cross without a flashlight, and it?s only strong enough to support two people at any given time. Each of the campers walks at a different speed. One can cross the bridge in 1 minute, another in 2 minutes, the third in 5 minutes, and the slow poke takes 10 minutes to cross. How do the campers make it across in 17 minutes?

Answer :

1 and 2 cross, taking 2 minutes, 1 goes back carrying the flashlight total = 3 minutes. 5 and 10 cross, taking 10 minutes totaltime now= 13 minutes, 2 goes back,total time now = 15 minutes. 1 and 2 cross again, taking 2 minutes making it 17 minutes.