Senior Software Engineer Interview Questions

Please go through the datastrucuture concepts

Hi... Could you please suggest some questions which was asked in your interview..?

no idea

What's the question?

Later tells me i am looking for few specific words

There are 5 Design Patterns: 1. The Factory Method - provides a simple decision making class that returns one of several possible subclasses of an abstract base class depending on the data that are provided. 2. The Abstract Factory Method - provides an interface to create and return one of several families of related objects. 3. The Builder Pattern - separates the construction of a complex object from its representation, so that several different representations can be created depending on the needs of the program. 4. The Prototype Pattern - starts with an initialized and instantiated class and copies or clones it to make new instances rather than creating new instances. 5. The Singleton Pattern - is a class of which there can be no more than one instance. It provides a single global point of access to that instance.

8*8 + 7*7 + 6*6 ... 1*1

total number of squares on chess board is given by recurrence relation: f(n) = f(n - 1) + n^2 f(8) = 7! + 8^2 = 140 + 64 so in total there is 204 squares... you can check this relation with smaller boards where its possible visually count squares..

Recruiter wanted to play chess instead of the software product domain level probabilities

it is 20-1 which is 19. Since 19 matches should be played to eliminate 19 people

It is n-1

1

use two stacks

Keep 2 stacks, let's call them inbox and outbox. Queue: - Push the new element onto inbox Dequeue: - If outbox is empty, refill it by popping each element from inbox and pushing it onto outbox - Pop and return the top element from outbox Using this method, each element will be in each stack exactly once - meaning each element will be pushed twice and popped twice, giving amortized constant time operations.

I was able to write the algorithm.

Your solution is really very helpful "I was able to write the algorithm" - congratulations, you got an "A+". However, were you afraid to divulge the solution because it may be proprietary or patentable? I, on the other hand have no such problems and following is my solution: void sort_01_arr(int *pArr, size) { int front = 0; //Sort array in ascending order, 0's first followed by 1's // Scan from the beginning looking for 1, when found scan from the end // looking for 0. If both found before crossing, swap the contents. // Continue until respective array indices (front and size) are equal or cross while(front< size) { if(pArr[front]) { // 1 if(pArr[size] == 0) { // 0 pArr[front++] = 0; pArr[size--] = 1; } else { --size; } } else { ++front; } } // elihw return; } Of course unarguably the simplest solution would be to have 2 passes through the array. In the first pass count how many 0's are present, and in the second pass just write that many 0' from the beginning of the array and then starting with the next element write 1's into the remaining elements.

its very easy that was asked in my last round.

Could you please elaborate answer.. I mean if you dont mind...Since i m not able to get the meaning of this que at first place,,