20 MCQ’s(Majority from OS and Java, C++, 1 aptitude)
2 coding questions.
1. Given an array, find the maximum sum that can be formed from the array such that no two adjacent elements are taken into consideration.
for ex:- 1,2,3,5 should return 7.
2. Print Vertical axis sum of the given binary tree.
Group Coding Round
36 people made it to the next round. 2 questions were asked
1. Given an array which initially increases and then decreases, search for an element in the array.
2. Merge ‘k’ sorted arrays.
18 people were selected for the next round and they asked me 2 questions
1. Given an infinitely growing sorted array which initially consists of 0’s and then 1’s upto infinity. Find the transition point where 0 changes to 1 effectively.
2. Given a binary search tree, make 2 separate trees such that difference between the sum of elements between them is minimum.(After some time he gave me a hint which helped me solve the problem).
1. Some basic OS related questions
2. Implement a data structure which would perform insertion, deletion, search and randomize operation with minimum time complexity.
We discussed a lot of data structures and I settled with a data structure (Hashtable with DLL). But he gave me clues and improvised my solution.
3. Implement a Queue using 2 stacks. Optimise your implementation
(They didn’t ask me to code in this round. They just checked how efficiently I approached it )
I think this is the bar raiser round. We are aware of the game show in which a contestant will think of a famous personality and the host will try to find the personality within a finite set of questions. I was asked to design a system which would implement the same. He asked me to write a code to return the first question (like the deciding factor on which the elements will be further partioned) so that I can find the celebritry in minimum no. of questions.
I think they were checking your approach, ideas and patience in this round. Whatever solution I gave he never seemed to be satisfied. Be confident in your approach and don’t ever give up.
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