Cognizant interview experience Real time questions & tips from candidates to crack your interview
SDE - 1
Cognizant
2 rounds | 2 Coding
problems
Selected Interview preparation journey
Preparation
Duration: 4 Months
Topics: Quant, Logical , DBMS, Data Structure, C , C++ language, HTML, CSS,SQL, Java
Tip 1 : You need a strong aptitude for knowledge as a first tip
Tip 2 : Solve questions from previous coding questions
Tip 3 : Make sure your resume includes at least two projects
Application process
Where: Campus
Eligibility: Above 6 cgpa
Tip 1 : Have at least 2 good projects mentioned in your resume with a link
Tip 2 : Focus on skills, internships, projects, and experiences.
Tip 3 : Make it simple, crisp, and one page
Interview rounds
01
Round
Easy
Online Coding Interview
Duration100 Minutes
Interview date19 Jun 2020
Coding problem1
Round1 is an Aptitude round that is on the Amcat platform. This round consists of four sections namely.
Quant (24 questions 35 minutes)
Logical (25 questions 35 minutes)
Code debugging (7 questions 20 minutes)
Essay question for 15 minutes.
1. Operating System Question
1. In the case of the index allocation scheme of various blocks to a file, the maximum size (possible) of the file would depend on :
a. the total number of blocks that have been used for the index, size of all the blocks
b. the actual size of all blocks, the size of the blocks’ address
c. the of the blocks’ size, the blocks’ address size, and the total number of blocks that have been used for the index
d. None of the above
Problem approach
Tip 1: Answer - A
Tip 2: Read Galvin for OS thoroughly.
02
Round
Easy
Video Call
Duration45 minutes
Interview date28 Jun 2019
Coding problem1
It was a technical Interview, interviewers were very friendly. There was one coding question from strings. And they have asked about the projects which I mentioned in my resume.
1. Tell me About Your Minor and Major project.
2. Can You Elaborate Major Project?
3. What is sllegalservices.in/ (A Personal Website created by me.)
4. Did You hosted in alone?
5. Tell me entire Process of Web Hosting?
6. Why don’t you hosted your website on cloud platform like AWS and Azure.
7. Are you aware about Python?
8. What is Django?
9. Enlist the Technologies used in this website.
10. What is Bootstrap?
11. What is PHP and MySQL?
12. Write Syntax to update column properties.
13. How will you combine two languages?
14. Write the code in chat box to combine two languages.
15. Give me the steps to create this Query sheet form (A form used in my website)
16. How will you retrieve the values submitted by this form?
17. Why you use only RDBMS in your Website?
18. Tell me the OSI layers?
19. In which layer is HTML in OSI model.
20. Tell me basic pillars of oops technology?
21. What is the main difference between JAVA and CPP.
1. Divide Two Integers
Easy
10m average time
90% success
0/40
Asked in companies
You are given two integers, ‘dividend’ and ‘divisor’.
You are required to divide the integers without using multiplication, division, and modular operators.
Your task is to return the quotient after dividing ‘dividend’ by ‘divisor’.
Note :
In cases where the dividend is not perfectly divisible by the divisor, you are required to return the integer value of the quotient which is closer to zero.
For example - If the answer is '8.34', we return the integer part, i.e., '8'. Also, If the answer is '-2.67', we return the integer part, i.e., '-2'.
Assume we're dealing with a system that can only store integers in the 32-bit signed integer range: [2^31, 2^31-1]. If the quotient is higher than 2^31 - 1, return 2^31 - 1; if it is less than -2^31, return -2^31.
For example :
If the answer is ‘9.333’, then return ‘9’, or if the answer is ‘-8.123’, then return ‘-8’.
Problem approach
This can be solved using bit manipulation using left shift .When we apply left shift operator to an integer it gets multiplied by 2. So I keep divisor multiplying by 2 until it become greater than the dividend and then again one time divide it by 2. After that i subtract divisor from dividend. and make the subtracted number divisor and apply the same process again until the dividend becomes less than divisor.
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