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I applied via Indeed and was interviewed before Aug 2022. There were 3 interview rounds.
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I applied via Referral and was interviewed in Feb 2020. There were 6 interview rounds.
I applied via Referral
I applied via Campus Placement
Prepare well for coding question based on leetocde medium level and hard level.try to understand code you write , they ask questions from coding round also
I am a software engineer with 5 years of experience in developing web applications using Java, Spring Boot, and Angular.
5 years of experience in software development
Proficient in Java, Spring Boot, and Angular
Developed web applications
I applied via Campus Placement and was interviewed in Sep 2024. There were 2 interview rounds.
50 questions was and the duration was 45 mins
Dsa and development questions were asked, and computer fundamentals were also asked.
I applied via Company Website and was interviewed before Jul 2022. There were 3 interview rounds.
Focus on Data structure problems and back tracking
It includes C programming questions and other language too
I appeared for an interview in Aug 2024.
A transaction in DBMS is a unit of work that must be executed as a whole. ACID properties ensure data integrity. SQL is relational, NoSQL is non-relational.
A transaction in DBMS is a set of operations that must be executed as a single unit.
ACID properties (Atomicity, Consistency, Isolation, Durability) ensure data integrity in transactions.
SQL databases are relational and use structured query language, while NoSQL data...
Encapsulation is the concept of bundling data and methods that operate on the data into a single unit. Inheritance allows a class to inherit attributes and methods from another class. Polymorphism allows objects of different classes to be treated as objects of a common superclass.
Encapsulation helps in data hiding and abstraction
Example: A class 'Car' encapsulates data like make, model, and methods like start(), stop()
...
Use Floyd's Tortoise and Hare algorithm to detect a cycle in a linked list.
Start with two pointers, slow and fast, moving at different speeds.
If there is a cycle, the two pointers will eventually meet at some point.
If there is no cycle, the fast pointer will reach the end of the list.
Example: 1->2->3->4->5->2 (cycle at node 2), slow and fast pointers will meet at node 2.
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