Code360 powered by Coding Ninjas X Naukri.com. Code360 powered by Coding Ninjas X Naukri.com

24/7 customer private limited interview experience Real time questions & tips from candidates to crack your interview

SDE - 2

24/7 customer private limited

2 rounds | 3 Coding problems

Anonymous

Experienced | Jul 2015

Rejected

Interview preparation journey

Preparation

Duration: 6 months

Topics: Data Structures, Algorithms, System Design, Aptitude, OOPS

Tip

Tip 1 : Must do Previously asked Interview as well as Online Test Questions.
Tip 2 : Go through all the previous interview experiences from Codestudio and Leetcode.
Tip 3 : Do at-least 2 good projects and you must know every bit of them.

Application process

Where: Other

Eligibility: Above 1 years of experience

Resume tip

Tip 1 : Have at-least 2 good projects explained in short with all important points covered.
Tip 2 : Every skill must be mentioned.
Tip 3 : Focus on skills, projects and experiences more.

Interview rounds

Round

Medium

Face to Face

Duration60 minutes

Interview date13 Jul 2015

Coding problem2

Technical Interview round with questions on DSA.

1. LRU Cache Implementation

Moderate

25m average time

65% success

0/80

Asked in companies

Design and implement a data structure for Least Recently Used (LRU) cache to support the following operations:

1. get(key) - Return the value of the key if the key exists in the cache, otherwise return -1.

2. put(key, value), Insert the value in the cache if the key is not already present or update the value of the given key if the key is already present. When the cache reaches its capacity, it should invalidate the least recently used item before inserting the new item.

You will be given ‘Q’ queries. Each query will belong to one of these two types:

Type 0: for get(key) operation.
Type 1: for put(key, value) operation.

Note :

1. The cache is initialized with a capacity (the maximum number of unique keys it can hold at a time).

2. Access to an item or key is defined as a get or a put operation on the key. The least recently used key is the one with the oldest access time.

Problem approach

Structure of an LRU Cache :

1) In practice, LRU cache is a kind of Queue — if an element is reaccessed, it goes to the end of the eviction order
2) This queue will have a specific capacity as the cache has a limited size. Whenever a new element is brought in, it is added at the head of the queue. When eviction happens, it happens from the tail of the queue.
3) Hitting data in the cache must be done in constant time, which isn't possible in Queue! But, it is possible with HashMap data structure
4) Removal of the least recently used element must be done in constant time, which means for the implementation of Queue, we'll use DoublyLinkedList instead of SingleLinkedList or an array.

LRU Algorithm :

The LRU algorithm is pretty easy! If the key is present in HashMap, it's a cache hit; else, it's a cache miss.

We'll follow two steps after a cache miss occurs:

1) Add a new element in front of the list.
2) Add a new entry in HashMap and refer to the head of the list.

And, we'll do two steps after a cache hit:

1) Remove the hit element and add it in front of the list.
2) Update HashMap with a new reference to the front of the list.

Try solving now

2. Minimum Number of Platforms

Moderate

30m average time

70% success

0/80

Asked in companies

You have been given two arrays, 'AT' and 'DT', representing the arrival and departure times of all trains that reach a railway station.

Your task is to find the minimum number of platforms required for the railway station so that no train needs to wait.

Note :

1. Every train will depart on the same day and the departure time will always be greater than the arrival time. For example, A train with arrival time 2240 and departure time 1930 is not possible.

2. Time will be given in 24H format and colons will be omitted for convenience. For example, 9:05AM will be given as "905", or 9:10PM will be given as "2110".

3. Also, there will be no leading zeroes in the given times. For example, 12:10AM will be given as “10” and not as “0010”.

Problem approach

The idea is to use the sorted arrival and departure times of trains and count the total number of platforms needed at a time . Maintain a counter to count the total number of trains present at the station at any point.
If the train is scheduled to arrive next, increase the counter by one and update the minimum platforms needed if the count is more than the minimum platforms needed so far.
If the train is scheduled to depart next, decrease the counter by 1.
Time Complexity : O(nlogn)

Try solving now

Round

Medium

Face to Face

Duration45 minutes

Interview date13 Jul 2015

Coding problem1

Technical Interview round with DSA based questions.

1. Reverse a linked list

Easy

15m average time

85% success

0/40

Asked in companies

You are given a Singly Linked List of integers. You need to reverse the Linked List by changing the links between nodes.

Example:

Input:
2 4 5 -1

Output:
5 4 2 -1

Explanation: 2->4->5 is the initial linked list. If we reverse this, we get 5->4->2.

Problem approach

This can be solved both: recursively and iteratively.
The recursive approach is more intuitive. First reverse all the nodes after head. Then we need to set head to be the final node in the reversed list. We simply set its next node in the original list (head -> next) to point to it and sets its next to NULL. The recursive approach has a O(N) time complexity and auxiliary space complexity.
For solving the question is constant auxiliary space, iterative approach can be used. We maintain 3 pointers, current, next and previous, abbreviated as cur, n and prev respectively. All the events occur in a chain.
1. Assign prev=NULL, cur=head .
2. Next, repeat the below steps until no node is left to reverse:
1. Initialize n to be the node after cur. i.e. (n=cur->next)
2. Then make cur->next point to prev (next node pointer).
3. Then make prev now point to the cur node.
4. At last move cur also one node ahead to n.
The prev pointer will be the last non null node and hence the answer.

Try solving now