Microsoft IDC Interview experience

ONLINE ASSESSMENT

Platform: CoCubes

Format: 3 coding questions

Time: 90 minutes

Q1) complete the following function:

int findMax(Treenode arr[], int size_of_array){ // code goes here }

Where Treenode is a structure defined as:

struct Treenode{ int feet; int inches; };

The function should calculate (12*feet+inches) for all the array elements and return the maximum value.

Q2) complete the following function:

Treenode* findInorderSuccessor( Treenode * root ,Treenode* node){ // code goes here } Where Treenode is a structure defined as: struct Treenode{ Treenode* left; Treenode* right; Int data; Treenode*parent; };

The function should return the pointer to the inorder successor of the ‘node’ provided in the function. If does not exist, return NULL.

Q3) complete the following function:

Node * findIntersection( Node* head1, Node*head2){ // code goes here } Where node is the structure of a linked list node defined as: struct Node{ int data; Node *next; };

Return the head pointer to the intersection of the two linked lists. It was mentioned that no extra space to be used and implementation should be recursive.

GROUP FLY ROUND:

40 candidates were shortlisted for the group fly round.

The candidates were roughly divided in a group of 4-5 and a mentor was in charge of each group. We were given a question and we were asked to write a function-solution in any high level language (scripting languages like Ruby, php, python were not allowed). We were allotted maximum 45 minutes.

Ques: Given two character arrays(not strings) of same length, and their length as a parameter to the function. We have to find whether the first string is a rotation of the other. We should not use any extra space. The time complexity may be quadratic.

The mentor kept coming to everyone. He first asked what i was thinking. I told him the approach with an example. Then I wrote the code, sample test cases and a basic approach section in bullets. Though, they demanded only the code.

**ROUND 1: **

Around 50% candidates were selected after the group fly round.

I was given two coding questions:

Ques 1) given a tree, print all the edges of the tree such that both the following conditions are satisfied:

Only print the node whose right child is NULL,

The node is a leaf node.

I gave a recursive approach. I later realised while explaining that it failed in certain conditions. I asked for some time to rectify it. I finally gave him a code and he seemed fine about it.

Ques 2) given a matrix, traverse the matrix in a zig-zag manner:

Ex: 1 2 3 4 5 6 7 8 9 1 1 2 Traversal: 1 2 5 9 6 3 4 7 1 1 8 2

I gave an approach of time complexity O(N X M). He asked me if i can do it in lesser time complexity. I may use extra space. I could not come up with anything very concrete. My round was done.

**ROUND 2:**

The interviewer asked how was my group fly round. Then he explained to me a situation that I have a fixed 2-D space of NXM. I have been given a set of random numbers(not in running). I have to find an efficient way to store them and such that I can retrieve each element in least time complexity.

With hit and trials we reached an approach where I can store the numbers in such a way that each row is sorted, and the order across the rows is also increasing. Now we can just apply binary search on 1st column to find the appropriate row , and then binary search the number in the obtained row. Time complexity: logM+logN. He asked me to code it. That was it for the second round.

The interviewer was very helpful and jolly.

ROUND 3: (FINAL)

I was called for round 3 after 15 minutes. It was supposed to be HR but it was predominantly technical. I was asked about all my projects and my role in them.

Then he gave me a question and said that we would discuss only the approach ,and there was no need to code it.

The problem was similar to the below geeksforgeeks problem:https://www.geeksforgeeks.org/divide-and-conquer-set-7-the-skyline-problem/

I struggled a bit but he helped me. Then we discussed the final solution and approach.

Since it was 8:30 pm, I was done for the day. After the round was over, I was told that I am done with all the rounds and need not come the other day.

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