Tip 1 : Practice at least 350 coding question
Tip 2 : Having recent project
Tip 3 : Good knowledge of OS,DBMS
Tip 1 : Have some projects on resume.
Tip 2 : Do not put false things on resume.
This was the initial round consisting of some aptitude questions,technical multiple choice questions and one coding question.
The Multiple Choice Questions were pretty straight forward.



For a (6 x 6) board, the numbers are written as follows:

You start from square 1 of the board (which is always in the last row and first column). On each square say 'X', you can throw a dice which can have six outcomes and you have total control over the outcome of dice throw and you want to find out the minimum number of throws required to reach the last cell.
Some of the squares contain Snakes and Ladders, and these are possibilities of a throw at square 'X':
You choose a destination square 'S' with number 'X+1', 'X+2', 'X+3', 'X+4', 'X+5', or 'X+6', provided this number is <= N*N.
If 'S' has a snake or ladder, you move to the destination of that snake or ladder. Otherwise, you move to S.
A board square on row 'i' and column 'j' has a "Snake or Ladder" if board[i][j] != -1. The destination of that snake or ladder is board[i][j].
You can only take a snake or ladder at most once per move: if the destination to a snake or ladder is the start of another snake or ladder, you do not continue moving - you have to ignore the snake or ladder present on that square.
For example, if the board is:
-1 1 -1
-1 -1 9
-1 4 -1
Let's say on the first move your destination square is 2 [at row 2, column 1], then you finish your first move at 4 [at row 1, column 2] because you do not continue moving to 9 [at row 0, column 0] by taking the ladder from 4.
A square can also have a Snake or Ladder which will end at the same cell.
For example, if the board is:
-1 3 -1
-1 5 -1
-1 -1 9
Here we can see Snake/Ladder on square 5 [at row 1, column 1] will end on the same square 5.
The idea is to consider the given snake and ladder board as a directed graph with number of vertices equal to the number of cells in the board. The problem reduces to finding the shortest path in a graph. Every vertex of the graph has an edge to next six vertices if next 6 vertices do not have a snake or ladder. If any of the next six vertices has a snake or ladder, then the edge from current vertex goes to the top of the ladder or tail of the snake. Since all edges are of equal weight, we can efficiently find shortest path using Breadth First Search of the graph.
Following is the implementation of the above idea. The input is represented by two things, first is ‘N’ which is number of cells in the given board, second is an array ‘move[0…N-1]’ of size N. An entry move[i] is -1 if there is no snake and no ladder from i, otherwise move[i] contains index of destination cell for the snake or the ladder at i.
1. Working of PING and TRACEROUTE command
2. OSI layers and it’s responsibility
3. Indexing in database
4. B/B+ tree used in index, asked me to explain node structure of both
5.Question based upon projects that I have mentioned in resume
What problems you faced and how you tackle them
Were you satisfied with outcome etc..
I had two interviewers in this round.Some Problem solving questions were asked in this round.
our enemy challenges you to play Russian Roulette with a 6-cylinder pistol (meaning it has room for 6 bullets). He puts 2 bullets into the gun in consecutive slots, and leaves the next four slots blank. He spins the barrel and hands you the gun. You point the gun at yourself and pull the trigger. It doesn't go off. Your enemy tells you that you need to pull the trigger one more time, and that you can choose to either spin the barrel at random, or not, before pulling the trigger again. Spinning the barrel will position the barrel in a random position.
Assuming you'd like to live, should you spin the barrel or not before pulling the trigger again?
You are better off shooting again without spinning the barrel.
Given that the gun didn't fire the first time, it was pointing to one of the four empty slots. Because your enemy spun the cylinder randomly, it would have been pointing to any of these empty slots with equal probability. Three of these slots would not fire again after an additional trigger-pull, and one of them would. Thus, by not spinning the barrel, there is a 1/4 chance that pulling the trigger again would fire the gun.
Alternatively, if you spin the barrel, it will point to each of the 6 slots with equal probability. Because 2 of these 6 slots have bullets in them, there would be a 2/6 = 1/3 chance that the gun would fire after spinning the barrel.
Thus, you are better off not spinning the barrel.



An Anagram is a word or phrase formed by rearranging the letters of a different word or phrase. We can generalise this in string processing by saying that an anagram of a string is another string with the same quantity of each character in it, in any order.
String “eat” and “ate” are anagram to each other but string “buy” and “bye” are not.
Explain all the main Object Oriented programming Concepts
I was asked to explain about any of my projects
Start off by taking an example.Then explain all the concepts one by one along with the example.
Number Of Interviewers:2
I was asked two problem solving questions and I was asked to explain my projects.
Given two hourglass of 4 minutes and 7 minutes, the task is to measure 9 minutes.
At 0 minutes: Start both hourglasses at the same time.
At 4 minutes: 4 minutes hourglass runs out and flip it. 7 minutes hourglass is left with 3 minutes.
At 7 minutes: 4 minutes hourglass is left with 1 minute. 7 minutes hourglass runs out and flip it.
At 8 minutes: 4 minutes hourglass runs out and 7 is filled with 6 minutes and 1 minute on the other side. Flip it as the sand is left with 1 minute.
At 9 minutes: 7 minutes hourglass becomes empty from above side.
Hence we measured 9 minutes.
he game of Tic-Tac-Toe is being played between two players and it is in below state after six moves.
X | O |
X | |
O | O | X
Who will win the game, O or X?
O will win the game.
The 7th mark must be placed in square 5 which is the win situation for both X and O. Hence, the 6th mark must be placed in a line already containing two of the opponents marks. There are two such possibilities – the 6th mark would have been either O in square 7 or X in square 9.
As we know both the players are intelligent enough, the 6th mark could not be O in square 7. Instead, he would have placed O in square 5 and would have won.
Hence, the sixth mark must be X placed in square 9. And the seventh mark will be O. Thus O will win the game.
Some basic HR questions were asked.I had one interviewer.
Questions asked:-
1)Did I know what package the company was offering .
2)Why I am not pursuing Masters or any other higher studies.
3)If I wanted to know anything about the company.
4)Why Cisco?

Here's your problem of the day
Solving this problem will increase your chance to get selected in this company
To make an AI less repetitive in a long paragraph, you should increase: