BarrierBreak Solutions Associate Software Engineer Interview Questions and Answers

Easy applitute question with 1 coding question at the end

Q1. String builder vs string buffer
Ans. 
String builder is faster but not thread-safe, while string buffer is slower but thread-safe.
• String builder is recommended for single-threaded applications for better performance.

• String buffer is recommended for multi-threaded applications to ensure thread safety.

• Example: StringBuilder sb = new StringBuilder(); StringBuffer sbf = new StringBuffer();
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Q2. Final keyword in oops
Ans. 
The final keyword in OOP is used to restrict the user from changing the value of a variable, overriding a method, or inheriting from a class.
• Final variable: value cannot be changed once assigned

• Final method: cannot be overridden in child classes

• Final class: cannot be inherited by other classes
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Q3. Bsf graph explanation
Ans. 
BSF graph explanation
• BFS (Breadth-First Search) is a graph traversal algorithm that explores all the neighbor nodes at the present depth before moving on to the nodes at the next depth.

• It uses a queue data structure to keep track of the nodes to visit next.

• BFS is often used to find the shortest path in an unweighted graph.

• Example: BFS can be used to find the shortest path in a maze from the starting point to the exit.
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It contains 60 questions solve in 60 minutes and 1 coding questions simply debug extra 15 minutes are given

Q1. 7 to 8 minut google meet technical interview about your project, sql and basics of your main language
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Q1. It just normal interview
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There was an OA which comprised of Aptitude questions, as well as Technical questions. And, there was a single coding question(Basic level)

Q1. This round was with the CTO of the company and manager, and CHRO of the company. Questions related to project were asked.
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Q2. Tell me something that is not there in your resume. Describe any one of your project in detail What is S3 architecture? Difference between GET and POST Describe HTTP Response
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It was online assessment

Q1. Questions were mainly from oops c++ and database management
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Q2. What is four pillars of oops. What is array and vector. Implement queue using stack. Project based questions.
Ans. 
The four pillars of OOP are encapsulation, inheritance, polymorphism, and abstraction. Arrays are fixed-size data structures, while vectors are dynamic arrays. A queue can be implemented using two stacks.
• Encapsulation: Bundling data and methods that operate on the data into a single unit.

• Inheritance: Creating new classes based on existing classes, inheriting their attributes and methods.

• Polymorphism: Objects of differe...read more
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Q3. Four pillars of opps
Ans. 
The four pillars of object-oriented programming are encapsulation, inheritance, polymorphism, and abstraction.
• Encapsulation: Bundling data and methods that operate on the data into a single unit.

• Inheritance: Allowing a new class to inherit properties and behavior from an existing class.

• Polymorphism: The ability for objects of different classes to respond to the same message.

• Abstraction: Hiding the complex implementatio...read more
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Q4. What is dbms explain
Ans. 
DBMS stands for Database Management System. It is a software system that allows users to define, create, maintain and control access to databases.
• DBMS is a software system that manages databases.

• It allows users to define, create, maintain, and control access to databases.

• DBMS provides tools for data manipulation, retrieval, and security.

• Examples of DBMS include MySQL, Oracle, SQL Server, and PostgreSQL.
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Q5. What isSQL.and some queries
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Easy level questions

Q1. Difference between object and constructor?
Ans. 
Object is an instance of a class with its own state and behavior, while constructor is a special method used to initialize objects.
• Object is created from a class using the 'new' keyword.

• Constructor is a special method with the same name as the class.

• Object has its own state (attributes) and behavior (methods).

• Constructor is used to initialize the state of an object when it is created.

• Example: Class Car { constructor(ma...read more
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Q2. Difference between abstraction and encapsulation?
Ans. 
Abstraction focuses on hiding the implementation details while encapsulation focuses on bundling the data and methods that operate on the data into a single unit.
• Abstraction is the concept of hiding the complex implementation details and showing only the necessary features of an object.

• Encapsulation is the bundling of data and methods that operate on the data into a single unit.

• Abstraction allows for creating simple in...read more
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Q1. Easy question on arrays
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Q2. Easy question on arrays and trees
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Simple aptitude questions, quantitative and qualitative questions

1 medium coding question. Preferably write in Java. Add commands to explain your code along with time and space complexity

Q1. Interview round where ask about general DSA questions, SQL along with practical questions.
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Q2. A variant of bubble sort
Ans. 
Cocktail sort is a variant of bubble sort that sorts in both directions.
• Also known as bidirectional bubble sort or shaker sort.

• It starts with the first element and compares it with the next one, swapping if necessary.

• Then it moves to the last element and compares it with the previous one, swapping if necessary.

• It continues until no more swaps are needed.

• It is more efficient than bubble sort for large lists with many ou
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Round duration - 50 minutes
Round difficulty - Medium

It was virtual 1st round with aptitude and computer science related questions and the other was the programming question for which they were given 30 min to solve and available languages are C,C++,Java and Python.

Q1. 
Ninja and Alternating Largest Problem Statement
Ninja is given a sequence of numbers and needs to rearrange them so that every second element is greater than its neighbors on both sides.
Example:
Input:
...read more
Ans. 
The task is to rearrange the given array such that every second element is greater than its left and right element.
• Read the number of test cases

• For each test case, read the number of elements in the array and the array elements

• Iterate through the array and swap elements at odd indices with their adjacent elements if necessary

• Check if the rearranged array satisfies the conditions and print 1 if it does, else print 0
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Round duration - 60 minutes
Round difficulty - Medium

It was around 1PM when this round started. It was taken by 2 developers working in the company. Initially I was a bit nervous but both ther interviewers are flexible. It went well with some discussions and questions.

Q1. 
Longest Common Prefix After Rotation
You are given two strings 'A' and 'B'. While string 'A' is constant, you may apply any number of left shift operations to string 'B'.
Explanation:
Your task is to calcu...read more
Ans. 
The question asks to find the minimum number of left shift operations required to obtain the longest common prefix of two given strings.
• Perform left shift operations on string B to find the longest common prefix with string A

• Count the number of left shift operations required to obtain the longest common prefix

• Return the minimum number of left shift operations for each test case
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Round duration - 30 minutes
Round difficulty - Medium

This was the final and technical round. It happened in between 7-7.30 pm. This interview was taken by senior member in the company.

Q1. 
Sub Sort Problem Statement
You are given an integer array ARR. Determine the length of the shortest contiguous subarray which, when sorted in ascending order, results in the entire array being sorted in a...read more
Ans. 
The question asks to find the length of the shortest contiguous subarray that needs to be sorted in order to sort the whole array.
• Iterate through the array and find the first element that is out of order from the left side.

• Iterate through the array and find the first element that is out of order from the right side.

• Find the minimum and maximum element within the subarray from the above steps.

• Expand the subarray from bo...read more
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Q1. Implent Oops concept of Java through code
Ans. 
Java OOPs concepts can be implemented using classes, objects, inheritance, polymorphism, and encapsulation.
• Create classes with attributes and methods

• Instantiate objects from classes

• Use inheritance to create subclasses

• Implement polymorphism through method overriding

• Encapsulate data using access modifiers
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