# Expedia Interview Experience(On Campus for internship 2020)

Expedia visited our campus in August for recruiting interns. The process began with an online round followed by 3 interview rounds.

The online round consisted of two sections: 2 coding questions and 10 technical multiple choice questions. Time limit: 90 minutes. Platform: HackerRank. This was a fairly easy round.

Q1. You are given a number n. You have to tell how many prime numbers less than or equal to n exist. For ex. if n=10, your answer should be 4 as {2, 3, 5, 7} are primes below 10.

Q2. You are given two strings A and B. You have to output a string C, which is formed by letterwise interleaving of A and B. For ex. if A=”abc” and B=”efg” output C should be “aebfcg”. Another ex. A=”abc” B=”pqrst” then C=”apbqcrst”.

About 25 students were selected for the next round.

**Round 1: Technical Interview 1**

Firstly, I was asked to introduce myself briefly. Then I was asked about my hobbies.

Next the interviewer asked me about my favourite data structure. I said Linked lists. So she gave me the following problems:

1.Write a program to reverse a linked list.

2. Write a program to reverse linked list in groups of given size ‘K’.

3.You are given a very large number which is in the form of a singly linked list such that each digit is a node. For ex. 123 is written as 1–>2–>3. Now you have to add 1 to the number and return its head.Remember you only have a singly linked list. So i told her that we can initially reverse the list.Alongside we maintain a variable carry and add 1 to the new head of the reversed linked list(that would be the last digit of the number). New value of the first node becomes (1+(initial value))%10 and carry for that node has the value (1+(initial value))/10. Also for all other nodes, new value=((carry+(initial value))%10 and carry=(carry+(initial value))/10 .

for ex. if we are given 129, so initially 1–>2–>9 becomes 9–>2–>1.First adding 1 to 9 gives us 10 so new value =10%10=0 and carry=10/10=1, so list changes to 0–>2–>1 with carry of 1 st node=1. New value of 2nd node=(1+2)%10=3 and carry=3/10=0., so list changes to 0–>3–>1 with carry of 2nd node=0. New value of 3rd node=(0+0)%10=0 and carry=0. so list becomes 0–>3–>1. Reversing the list we get 1–>3–>0 which is required answer.

She was satisfied with my solution and then asked for any special cases for this problem. I thought for a while and responded that if initially we are given a number with all 9’s as digits we would have to add an extra node(with value 1) in the resultant list to get the answer. For ex. we are given 9–>9–>9 reversing it gives us 9–>9–>9 we add 1 to head alonwith the carry method which gives us 0–>0–>0 and 1 carry over left. So we add an extra node with value 1, i.e. 0–>0–>0–>1 and reverse it to get 1–>0–>0–>0.

13 students were further selected.

**Round 2: Technical Interview 2**

Again, I was asked to introduce myself briefly. Then they asked me questions based on my CV.

After that came the questions:

- How to design a tiny URL shortener. Solution: https://www.geeksforgeeks.org/how-to-design-a-tiny-url-or-url-shortener/
- Given two linked lists, find if they are merging at some node or not. One solution I gave was that for the first linked list we traverse it completely and store the references in a hashset(Java). Now while traversing the second linked list if we find a node whose reference is already stored in our hashset, we have found the merging node else no merging node exists. But since this method used memory I was asked to optimise it. So I proposed another solution: We can find the difference in sizes of the 2 lists (D), Traverse the longer list by D nodes and then simultaneously traverse both lists together, if at any point we find that list1.next==list2.next, We have found the merging node else no such node exists.
- Find the length of the longest Consecutive 1s in Binary Representation of a number. For ex. n=14 Binary representation=1110 which should give us output as 3. Refer to:https://www.geeksforgeeks.org/length-longest-consecutive-1s-binary-representation/

10 students were selected for the final round.

**Round 3: HR round**

Unfortunately, I was not selected for the HR Round.

But the questions asked in the HR round were quite general like tell us where you see yourself in 5 years, why do you want to join us etc.

Finally 6 students got selected.

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