Credit Suisse interview experience Real time questions & tips from candidates to crack your interview

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Credit Suisse
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3 rounds | 7 Coding problems

Interview preparation journey

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Preparation
Duration: 5 months
Topics: Data Structures, Algorithms, DBMS, Aptitude, OOPS
Tip
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: Campus
Eligibility: Above 7 CGPA
Resume Tip
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

01
Round
Easy
Coding Test - Pen and paper
Duration90 minutes
Interview date4 Jul 2016
Coding problem1

The test was a written test with no negative marking!
15 Quantitative analysis and reasoning questions (MCQs). The problems were basic to mid-level CAT. 
Quantitative 7 Puzzles (Subjective)-How to Ace the Brain Teasers book helps
5 Programming questions (Subjective)Algorithm or code in any language was accepted.-Questions were based on String manipulation techniques, Searching-Sorting techniques, anagram. 
1 Bonus question (Subjective)-Client Server Architecture type of question.

1. Angle Between Hour Hand And Minute Hand

Easy
15m average time
85% success
0/40
Asked in companies
MicrosoftSalesforceAmazon

Given the time in hours and minutes, you need to calculate the angle between the hour hand and the minute hand.

Note :
There can be two angles between the hour hand and minute hand, you need to print a minimum of two. Also, print the floor value of angle i.e. if the angle is 15.2, you need to print 15.
Problem approach

The idea is to take 12:00 (h = 12, m = 0) as a reference. 
Steps : 
1. Calculate the angle made by hour hand with respect to 12:00 in h hours and m minutes. 
2. Calculate the angle made by minute hand with respect to 12:00 in h hours and m minutes. 
3. The difference between the two angles is the angle between the two hands

Pseudocode :
calcAngle(h,m)
{
// validate the input
if (h <0 or m < 0 or h >12 or m > 60)
print("Wrong Input")

if (h is equal to 12) h = 0;
if (m is equal to 60)
{
m = 0;
h += 1;
if(h>12)
h = h-12;
}

// Calculate the angles moved by hour and minute hands with reference to 12:00
initialise hour_angle as 0.5 * (h * 60 + m)
initialise minute_angle as 6 * m

// Find the difference between two angles
initialise angle as absolute value of (hour_angle - minute_angle)

// Return the smaller angle of two possible angles
update angle to minimum of (360 - angle, angle)

return angle 
}

Try solving now
02
Round
Easy
Face to Face
Duration60 minutes
Interview date4 Jul 2016
Coding problem5

Technical round with questions on OOPS/DBMS etc.

1. DBMS Question

Types of Indexing

Problem approach

Indexing in Database is defined based on its indexing attributes. Two main types of indexing methods are:
1. Primary Indexing : Primary Index is an ordered file which is fixed length size with two fields. The first field is the same a primary key and second, filed is pointed to that specific data block. In the primary Index, there is always one to one relationship between the entries in the index table.
The primary Indexing in DBMS is also further divided into two types.
Dense Index
Sparse Index

2. Secondary Indexing : The secondary Index in DBMS can be generated by a field which has a unique value for each record, and it should be a candidate key. It is also known as a non-clustering index.
This two-level database indexing technique is used to reduce the mapping size of the first level. For the first level, a large range of numbers is selected because of this; the mapping size always remains small.

2. DBMS Question

Types of Keys in DBMS

Problem approach

Super Key – A super key is a group of single or multiple keys which identifies rows in a table.
Primary Key – is a column or group of columns in a table that uniquely identify every row in that table.
Candidate Key – is a set of attributes that uniquely identify tuples in a table. Candidate Key is a super key with no repeated attributes.
Alternate Key – is a column or group of columns in a table that uniquely identify every row in that table.
Foreign Key – is a column that creates a relationship between two tables. The purpose of Foreign keys is to maintain data integrity and allow navigation between two different instances of an entity.
Compound Key – has two or more attributes that allow you to uniquely recognize a specific record. It is possible that each column may not be unique by itself within the database.
Composite Key – is a combination of two or more columns that uniquely identify rows in a table. The combination of columns guarantees uniqueness, though individual uniqueness is not guaranteed.
Surrogate Key – An artificial key which aims to uniquely identify each record is called a surrogate key. These kind of key are unique because they are created when you don’t have any natural primary key.

3. DBMS Question

Mapping ER diagram to relational schema

Problem approach

Following rules are used for converting an ER diagram into the tables-

Rule-01: For Strong Entity Set With Only Simple Attributes-
A strong entity set with only simple attributes will require only one table in relational model.
Attributes of the table will be the attributes of the entity set.
The primary key of the table will be the key attribute of the entity set.

Rule-02: For Strong Entity Set With Composite Attributes-
A strong entity set with any number of composite attributes will require only one table in relational model.
While conversion, simple attributes of the composite attributes are taken into account and not the composite attribute itself.

Rule-03: For Strong Entity Set With Multi Valued Attributes-
A strong entity set with any number of multi valued attributes will require two tables in relational model.
One table will contain all the simple attributes with the primary key.
Other table will contain the primary key and all the multi valued attributes.

Rule-04: Translating Relationship Set into a Table-
A relationship set will require one table in the relational model.
Attributes of the table are-
Primary key attributes of the participating entity sets
Its own descriptive attributes if any.
Set of non-descriptive attributes will be the primary key.

Rule-05: For Binary Relationships With Cardinality Ratios-
The following four cases are possible-
Case-01: Binary relationship with cardinality ratio m:n
Case-02: Binary relationship with cardinality ratio 1:n
Case-03: Binary relationship with cardinality ratio m:1
Case-04: Binary relationship with cardinality ratio 1:1
While determining the minimum number of tables required for binary relationships with given cardinality ratios, following thumb rules must be kept in mind-
For binary relationship with cardinality ration m : n , separate and individual tables will be drawn for each entity set and relationship.
For binary relationship with cardinality ratio either m : 1 or 1 : n , always remember “many side will consume the relationship” i.e. a combined table will be drawn for many side entity set and relationship set.
For binary relationship with cardinality ratio 1 : 1 , two tables will be required. You can combine the relationship set with any one of the entity sets.

Rule-06: For Binary Relationship With Both Cardinality Constraints and Participation Constraints-
Cardinality constraints will be implemented as discussed in Rule-05.
Because of the total participation constraint, foreign key acquires NOT NULL constraint i.e. now foreign key can not be null.

Case-01: For Binary Relationship With Cardinality Constraint and Total Participation Constraint From One Side-
Because cardinality ratio = 1 : n , so we will combine the entity set B and relationship set R.

Case-02: For Binary Relationship With Cardinality Constraint and Total Participation Constraint From Both Sides-
If there is a key constraint from both the sides of an entity set with total participation, then that binary relationship is represented using only single table.

Rule-07: For Binary Relationship With Weak Entity Set-
Weak entity set always appears in association with identifying relationship with total participation constraint.

4. OOPS Question

Differences between Procedural and Object Oriented Programming

Problem approach

1. In procedural programming, program is divided into small parts called functions. In object oriented programming, program is divided into small parts called objects.
2. Procedural programming follows top down approach. Object oriented programming follows bottom up approach.
3. There is no access specifier in procedural programming. Object oriented programming have access specifiers like private, public, protected etc.
4. Adding new data and function is not easy in Procedural programming. Adding new data and function is easy in Object oriented programming.
5. Procedural programming does not have any proper way for hiding data so it is less secure. Object oriented programming provides data hiding so it is more secure.

5. OOPS Question

What is encapsulation, data hiding and abstraction?

Problem approach

Encapsulation binds the data & functions together which keeps both safe from outside interference. Data encapsulation led to data hiding.
Data Abstraction is a mechanism of hiding the implementation from the user & exposing the interface.
Data hiding is a process of combining data and functions into a single unit. The ideology behind data hiding is to conceal data within a class, to prevent its direct access from outside the class. It helps programmers to create classes with unique data sets and functions, avoiding unnecessary penetration from other program classes.

03
Round
Easy
HR Round
Duration30 minutes
Interview date4 Jul 2016
Coding problem1

HR round with typical behavioral problems.

1. Basic HR Questions

1) Tell me about yourself? What do you do in your leisure time?
2) Why Investment Banking?
3) One quality that makes me different from the rest of my class with example
4) How have I improved after internship interview experience with Credit Suisse?
5) Why no further studies in spite of good grades?
6) Why Credit Suisse?

Problem approach

Tip 1 : The cross questioning can go intense some time, think before you speak.
Tip 2 : Be open minded and answer whatever you are thinking, in these rounds I feel it is important to have opinion.
Tip 3 : Context of questions can be switched, pay attention to the details. It is okay to ask questions in these round, like what are the projects currently the company is investing, which team you are mentoring. How all is the work environment etc.

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Skill covered: Programming

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