HashedIn Interview Questions

• Round 1: Online Coding Test. It consists of 3 coding questions. Generally, the two questions belong to easy and medium. And one question belongs to hard. The test is conducted on the Codility platform.
• Round 2: Technical Round 1. This round was a DSA-based round in which the interviewer basically starts with your introduction and then moves to  2 coding questions. The questions generally belong to easy and medium levels and in some cases, they may be hard levels. In this, the interviewer wants you to explain the approach to the problem. If you explain Bruteforce then the interviewer wants you to optimize the code, and then you are required to write a rough code in Google docs along with the Asymptotic time complexity of the code. The duration of the round generally lasts from 1 to 1.5 hours. The call will happen over Zoom Meeting.
• Round 3: Technical Round 2. This round will be based on your Computer Fundamental skills, in which the interviewer can ask questions on the topics- Object-Oriented Programming,  Programming Language on Resume, Operating System Questions, DBMS Questions, and Computer Network Questions. And finally, the interviewer will ask System Design Questions. The duration of the round also lasts from 1 to 1.5 hours. The call will happen over Zoom Meeting.
• Round 4: Technical Round 3 (Optional). This round is optional. For some profiles, it is being conducted. This round is generally conducted to validate the feedback from the previous rounds. It consists of fundamental questions followed by two DSA-based questions. The interviewer expects the same as in Round 2. This round will be of a max duration of 1 hour. And the call will be on Zoom meeting.
• Round 5: Fitment Round. This final round consists of behavioral questions, About, Family Background, And Typical HR Questions. The duration of this round will be between 20 - 30 minutes over the zoom meeting. This round is generally to test your communication skills and also how well you can interact with people. Generally, the Interviewer starts with your basic information like - Introduction, Family Background, Education, and College. Then the interviewer may move on to some tricky questions that you need to answer well.
  • Why do you want to work for our company? 
    This question is an important one, especially if the job is for a startup. Startups are often run by a small team that has fewer resources than large corporations and even government agencies. There's less job security, too, which means you may be asked to do more without the same benefits as a full-time employee. Before answering this question, consider what the job is for and for whom. Is it for a company that is growing, or is it for an established brand that has a stable of products with loyal customers? This question can also be used to determine if you will be happy as part of the team.
  • Why do you want this job?
    This question is a little trickier than the first one, but it's also more important. Why do you want to work for a certain company? What is it about this particular job that you like? What skills from your past experiences would you like to use in your next job? These are all good questions to ask yourself when considering why you want to work for a company. While it is important to answer this question for your own sake, it is also an important question for the company. If the answer is unrelated to the job, then it will reveal why you want the job, and the employer may choose to not hire you.
  • What are your strengths?
    Another common interview question is the strength question. Here, the interviewer will want to know what qualities you have in abundance, and what you excel at. This will give them a sense of what your work environment will be like, and what challenges they may face while having you on their team. This can be used to judge if you will fit well in a particular position, but it can also serve as a good icebreaker that gets the conversation started.
  • What are your weaknesses?
    This is a tricky one, and you may even get some sideways glances if you say it out of turn. But it is an important question that can reveal some characteristics that you may not want to highlight before the job interview. A weakness question is used to see if you are honest and open about your shortcomings. If you are, then it means you are being genuine and honest with the interviewer, which is a good thing.
  • Can you demonstrate what you just talked about?
    You may want to avoid this question and the one that follows because it is where people often trip themselves up. If you have been asked to demonstrate something you may have said during your discussion, you may want to consider what you are demonstrating. If you have been asked to walk the interviewer through your strengths, then the right answer is a resounding yes. If, however, you have been asked to show off some skill, you may want to consider politely declining.
  • Could you give us an example of when you used each one of these skills?
    If the interviewer asks you to provide an example of a certain but, u, or v attribute you possess, then they want to see it in action. Think carefully about how you want to demonstrate these skills in a real-life scenario. Don't just say that you are able to think on your feet, for example: Show the interviewer how you were able to think about a scenario and formulate a response that is both professional and appropriate for the situation.

HashedIn recruits freshers from on-campus, off-campus, and Employee Referrals. For experienced people, hashedIn follows direct hiring from the career portal and from employee referrals as well. HashedIn follows a common recruitment process that consists of four Rounds:

• Round 1: Online Coding Test.
• Round 2: Technical Round 1.
• Round 3: Technical Round 2.
• Round 4: Technical Round 3 (Optional). 
• Round 5: Fitment Round.

Example - Output array = [5, 7, 7, 11, 11, 0] 

Explanation - 

5 is the next greater element of 4.

7 is the next greater element of 5.

7 is the next greater element of 3.

11 is the next greater element of 7.

11 the next greater element of 2.

Last is 0 because there is no next greater element to this.

Solution: 

The intuitive solution is to find the greater element just after the current element. For this, we can use the brute force approach by using the nested loop. But this solution's time complexity will be O(n^2). Which is not very optimal. So for optimization, we can use a monotonic stack.

import java.util.Stack;
import java.util.ArrayList;
import java.util.List;
public class InterviewBit
{
    public static List<Integer> nextGreaterElement(int[] arr){
        //Integer Array that stores the answer.
        List<Integer> ans = new ArrayList<>();
        
        //Using a stack for finding the solution.
    	Stack<Integer> stack = new Stack<>();
    	
    	for(int i = 0; i < arr.length; i++){
    	    //Pushing element onto the stack either stack is empty or element is less than stack top. 
            if(stack.isEmpty() || arr[i] < stack.peek()) {
        		stack.push(arr[i]);
        	}else{
        	    
        	    //If found the element greater than stack top then that is the next greater element so we need to add it to the answer list.
        		while(!stack.isEmpty() && stack.peek() < arr[i]){
        			ans.add(arr[i]);
        			stack.pop();
        	    }
        	    //Pushing that greater element in stack for next cycle
        		stack.push(arr[i]);
        	}
        }
        //Adding the 0 to the answer array if no greater element exists.
        while(!stack.isEmpty()){
        	ans.add(0);
        	stack.pop();
        }
        return ans;
    }
	public static void main(String[] args) {
		System.out.println(nextGreaterElement(new int[]{4,5,3,7,2,11}));
	}
} 

The time complexity of the above approach is O(n) because we are only visiting once in every element. Because we are using a stack, the space complexity will also be O(n).

Example -

Input: prices = [4,7,1,8,2,9]

Output: 9

Explanation: Buy stocks on day 3 (price = 1) and sell on day 6 (price = 9), profit = 9-1 = 9=8.

Solution:

Again the brute force way is to try out all the possibilities by purchasing the stock on each day and try selling it on each day next to it. And from that get the max profit we can get. The time complexity for this will be O(n^2). So to optimize this we try the dynamic programming approach by finding the maximum profit seen so far and what can be the current profit for each day. So the solution for this approach will be - 

public class InterviewBit
{
    public static int maxProfit(int[] prices) {
        //Variable to keeptrack of minimum price
        int minPrice = Integer.MAX_VALUE;
        
        //Variable to keep track of maximum profit seen so far.
        int profit = 0;
        for(int i = 0; i < prices.length; i++){
            minPrice = Math.min(minPrice, prices[i]);
            profit = Math.max(profit, (prices[i] - minPrice));
        }
        return profit;
    }

	public static void main(String[] args) {
		System.out.println(maxProfit(new int[]{4,7,1,8,2,9}));
	}
}

The time complexity of the above approach is O(n) because we are traversing to each element only once. And the space complexity is O(1) constant because it only keeps track of a constant variable for any set of input.

Explanation - 

From the right side view the direct node we can see is [8, 9, 3, 25]. The other nodes like 12 are hidden behind 3. So it cannot be printed.

Solution:

For Solving this problem, we can traverse the binary tree with the level order and we can print the last node of the level. Level order traversal or BFS (Breadth First Search). We know that we have to use a queue data structure for level breadth first search on trees. So the Solution function will be -

//This Function returns the list of nodes that are visible from the right side.
    //This function accepts the root node of the tree.
    public List<Integer> rightSideView(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        
        if(root == null)
            return ans;
        
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        
        while(!queue.isEmpty()){
            int n = queue.size();
            for(int i = 0; i < n; i++){
                TreeNode curr = queue.poll();
                //adding the node to the answer if it is the last node .
                if(i == n-1)
                    ans.add(curr.val);
                
                //Add the child node to the queue. 
                if(curr.left != null)
                    queue.add(curr.left);
                if(curr.right != null)
                    queue.add(curr.right);
            }
        }
        return ans;
    }

In the above function, the TreeNode structure will be like - 
public class TreeNode {
      int val;
      TreeNode left; //Pointer to the left-subtree.
      TreeNode right; //Pointer to the right-subtree.
 }

The above method returns the list containing the nodes that are visible from the right side of the tree. In the above code, we are traversing every node of the tree. So the time complexity will be O(n). And hence we are using the Queue for traversing the tree, the space complexity will also be O(n).

For example, if the string “34” is provided, then the possible solutions are - ["dg","dh","di","eg","eh","ei","fg","fh","fi"].

Solution

As a result, we need to generate all possible letters. So here we have to apply the brute force backtracking approach because we must explore all possible solutions. So the backtracking solution will be -

class InterviewBit {
    List<String> ans; //answer that stores all the combinations.
    
    //method that returns the possible letters from the given keys.
    private String letters(char ch){
        switch(ch){
            case '2': return "abc";
            case '3': return "def";
            case '4': return "ghi";
            case '5': return "jkl";
            case '6': return "mno";
            case '7': return "pqrs";
            case '8': return "tuv";
            case '9': return "wxyz";
            default: return "";
        }
    }
    //Method that recursively explores the path to generate the answer.
    private void solution(String digits, int i, StringBuffer sb){
    //Base case when letters end then the possible combination of letters are added to the answer list.        
        if(i == digits.length()){
            ans.add(sb.toString());
            return;
        }
        //Getting each character and generating all the combinations.
        String letter = letters(digits.charAt(i));
        for(int j = 0; j < letter.length(); j++){
            sb.append(letter.charAt(j));
            //Recursively generating the other possible combination of letters.
            solution(digits, i+1, new StringBuffer(sb));
            //Backtracking
            sb.deleteCharAt(sb.length()-1); 
        }
    }
    public List<String> combinationLetters(String digits) {
        ans = new ArrayList<>();
        solution(digits, 0, new StringBuffer());
        return ans;
    }
}

Explanation:

In the above figure you can see that the last node connected with node 2. Here the Cycle exists. So we need to return true, which means the list contains a loop.

Solution:

Although there can be many solutions to this problem, one is to start with every node and check if we are reaching the same node again. But this is not a very optimal solution. Another solution is to start traversing the list and keep track of every node in the HashMap while traversing and check if that node is already present in the HashMap. If the node is already present then that means there is a cycle, so we need to return true. This solution is optimal in terms of time, but it consumes extra space. So for avoiding space complexity, we can use the two-pointer approach of slow and fast pointers and check that if both slow and fast pointers reach the same node, then that means the list contains a cycle. So let’s implement this algorithm.

public boolean hasCycle(ListNode head) {
        //Edge case
        if(head == null || head.next == null)
            return false;
        
        ListNode slow = head, fast = head;
        
        //Fast moves 2 steps and slow moves 1 step. If there is a cycle then both pointers will meet and that means the cycle exists in the list.
        while(fast != null && fast.next != null){
            fast = fast.next.next;
            slow = slow.next;
            if(slow == fast)
                return true;
        }
        //No cycle found so return false
        return false;
    } 

The above algorithm takes O(n) time because all the nodes travel only once. But if there is a cycle then the nodes will be visited only for some constant time. And constant O(1) space because no extra memory is required to detect the loop.

For example -  Consider the below tree and target = 26. And we can see the path exists in the tree from 9 -> 7 -> 6 -> 4. This path to 26. So we need to return true.

Solution -  For the solution, we can traverse the one sub-tree using DFS. And once we found that we reached the leaf node and found the sum equal to the target node then we can return true. Otherwise, we need to traverse the other sub-tree for finding the target and leaf node.

//The method returns the boolean result if there is a path.
public boolean solution(TreeNode root, int targetSum){
        //Base case when we reach to the leaf node and reached the target sum
        if(root.left == null && root.right == null){
            if(targetSum-root.val == 0)
                return true;
            return false;
        }
        boolean left = false;
        //Search on the left sub-tree for the path.
        if(root.left != null){
            left = solution(root.left, targetSum-root.val);
        }
        //Searching on the right sub-tree if there is no path in left sub-tree
        if(!left){
            if(root.right != null)
                return solution(root.right, targetSum-root.val);
        }
        return left;
    }

In the above approach, every node in the worst case will be traversed. So the time complexity will be O(n). And we are using recursion, so the space complexity will be also O(n) in the worst case.

There are also many coding questions that are asked in HashedIn Interview. But the patterns that are followed are basically covered in the above questions.  

HashedIn follows the same round for the senior roles. And also the levels of questions are almost similar. The rounds that differentiate the Interview rounds for the senior job profile are the proficiency and experience in development. General Questions related to the java expertise and some popular frameworks are asked. Like Java 8 concepts, Spring boot, Hibernate, Kafka, HLD design of the current project, LLD of e-commerce website, etc.

If you already know about Spring architecture, then you’ll be happy to know that Spring Boot architecture is built on top of the original architecture. It’s designed to address common problems related to scalability and speed. A Spring Boot application is a collection of several Java classes that work together to perform a specific task. Ideally, there should be three types of classes in your application - services, controllers, and repositories. Services are responsible for storing data, retrieving data from APIs, or sending data to APIs. Controllers receive data from services and save them as models so they can be returned to the user. Repositories are responsible for storing and retrieving data from a database. Although the application architecture might vary from project to project, there are a few things that should always remain the same. These include things like where you save your files, what programming language you use, and how the files are organized.

As of right now, you can use Java’s new stream feature if you’re coding in Java 9 (or higher). You’ll need to check that the code you’re using is compatible with Java 9 if you want to use streams, but as its adoption rate increases, that will change. If you want to use stream functionality in your code now, you can use the new Java 9-compatible alternatives for the old Java 8 stream operations (such as the Collectors class for collecting data, or the Map/Filter/Map for filtering and mapping operations). However, the new stream features are expected to receive improvements and additional functionality over the next few years, so you’ll need to keep an eye out for any changes.

Java’s stream API is currently in its infancy and is expected to grow and evolve over time as more developers get their hands on it and provide feedback. While you can expect to see improvements as Java 9 is adopted by developers, you can also help shape the direction of these APIs by providing feedback. You can do this by keeping an eye on the Java bug tracking system and submitting any bugs you find, as well as providing feedback directly to Oracle. You can also participate in discussions with other Java developers on social media or on various coding forum websites.

Useful Resources

• Java Interview Questions
• SDET Interview Questions
• QA Interview Questions
• SQL Interview Questions
• DevOps Interview Questions
• Data Engineer Interview Questions
• Software Engineering Interview Questions
• Software Testing Interview Questions
• Technical Interview Questions
• Coding Interview Questions
• Interview Resources
• Java Compiler
• Mock Interview

Exception is the Unexpected situation that may happen in the program that breaks the normal flow of the program. An exception is usually caused by unexpected situations like invalid user input, network connection failure, disk full, and so on. An exception is often represented by the java keyword throw when raising an exception and try and catch keywords for handling it.

Context switching is often done by a kernel’s task scheduler, which maintains a list of processes that are currently executing and another list of processes that are ready to execute but not currently executing. When a process on the ready list needs to be switched to the executing list, either because it has completed its task or it has been given a new task, the kernel’s task scheduler determines which process on the executing list to switch out, and which process on the ready list to switch in.

Operating Systems (OS) manage and handle all hardware and software resources on a computer. In the early 1950s, GMOs, or general-purpose operating systems, were introduced. OS handles, coordinates, and manages overall activities and shares computer resources. An OS acts as an intermediary between users of computers and the hardware they use.

Functions of Operating System -

The OS performs many tasks. The important OS capabilities are listed below.

• OS manages the memory of the System and Processor.
• OS provides an Interface for users where users can interact with System.
• OS also handles File Management and Device Management.
• OS Schedules resources and jobs for the CPU.
• OS also helps in error detection.
• OS also manages the Security of the System.

VPN (Virtual Private Network) is the private network that establishes a secure connection between two nodes. 

For Example - The client’s computer is connected to the organization’s network via the internet. The VPN software helps protect the computer and the data being transmitted. The VPN client software creates a secure connection to a remote server. The VPN server acts as a go-between, receiving data from your computer and then forwarding it to the destination you desire — be it a website, remote network, or some other computer. VPNs work by putting your computer in a “tunnel” that blocks all uninvited guests. It does this by encrypting your data and routing it through a different IP address — one belonging to the VPN company. VPNs also keep you anonymous by not logging your IP address.

RAID (Redundant Arrays of Independent Disks) is an information virtualization technology that uses multiple hard disks to store data. It combines independent hard drives to store data in order to save data. It balances data security, system performance, storage capacity, and other variables. Data storage performance and reliability are enhanced with RAID. It improves the performance and reliability of data storage. RAID increases storage capacity and minimizes data loss by providing data redundancy.

The most common form of IPC is the use of shared memory to pass data between processes. Shared memory is a portion of system memory that can be read and written to by multiple processes. When a process tries to read from or write to a portion of shared memory, it receives OS approval. Once the OS gives its approval, the process gets a copy of the shared memory and can read from or write to it. Any other process that tries to read from or write to the shared memory receives OS approval and will receive its own copy of the shared memory.

Different IPC Mechanisms:

• Pipes - It is a communication channel between two or more related or interrelated processes. It may be within one process or communication between child and parent process.
• Message Queuing - It is a temporary memory area that is provided by underlying operating systems and used to facilitate the communication between processes.
• Semaphores - It protects shared resources that are accessed by many processes simultaneously such as global shared memory. It serves as a guard for resources.
• Socket - In a two-way communication system, the socket provides a point-point connection between two processes.
• Shared Memory - It is a storage area in which multiple processes can access a common pool of data. Changes made by one process can be viewed by another process.
• Signals - It is a notification sent to a process informing it of the occurrence of an event.

A semaphore can be used to synchronize access to critical sections in a program by using the two atomic operations wait() and signal(). The atomic operation signal() indicates that the critical section is available, and the atomic operation wait() prevents other processes from entering the critical section until the current process finishes using it. A semaphore can only hold one positive integer value. 

Semaphores are used in situations where you have to control the flow of activities and/or secure a resource. For example, let’s say you are working with a company and you have to create a system that will only allow a certain number of employees to log in at any given time. You can use semaphores to control the flow of employees logging in to the system and prevent the system from being overloaded. Semaphores are widely used in real-time systems like communications, real-time control systems, process control systems, manufacturing systems, and so on.

DHCP (Dynamic Host Configuration Protocol) is used in most enterprises as well as homes to facilitate the server-client model where the DHCP server is the server and all the devices on the network are the clients. The DHCP server sends a lease to the clients that they can use the IP address for a certain amount of time. After the lease expires, it gets a new IP address from the DHCP server. The client device can request a new IP address from the DHCP server when it has exceeded its lease. The DHCP server has a pool of IP addresses from which it allocates to the clients.

A database transaction is a set of operations that are treated as a single unit of work. If one operation fails, then the entire transaction is rolled back (canceled) and the data is restored to the state it was in before the transaction started. Transactions ensure that the database maintains its integrity by preventing data from being corrupted or lost if one of its components fails. If no transaction is specified, the database system will treat each SQL statement as an independent transaction, which can lead to inconsistent data and errors if any of the SQL statements fail. 

The main reason database transactions are important is to ensure data integrity. When using a database, there is always a risk that something might go wrong and cause an error, or worse, corrupt data. Transactions can be set to automatically roll back if an error occurs, thus preventing data from being changed in the case of an error. If you’re writing data to your database, it’s important to use database transactions. By doing so, you can ensure that your data is written correctly and that if there’s an error, it will be rolled back so no data is lost. If you don’t use database transactions, an error could cause your data to be written incorrectly or not written at all.

DBMS is a software application used to create, manage, and store databases. Some of the most popular database management systems are Microsoft SQL Server, Oracle, MySQL, and IBM DB2. The most important function of a database management system is to store and retrieve data from a database. The data can come from almost any source, from different kinds of sensors or scanners, or from a legacy system such as ERP or HRM. A database contains information that can be accessed, updated, and shared. It is stored on a server, accessed by many people at the same time, and can be used to analyze trends and make predictions. A database management system is a software used to manage the database: create it, populate it with data, and secure it.

When the firewall is enabled, it restricts the traffic between the private network and the public network. It also monitors the traffic to detect any malicious activity. A firewall can be software or hardware-based. It is widely used in offices, schools, and other organizations. It is a very important security system because it protects the data from hackers. It helps prevent cyber-attacks.

Deadlock is the problem that arises in the operating system while executing the processes. To understand this problem better, let’s see an example of two people, who are working on a construction site. Both are responsible for a certain task and they have to work simultaneously. But their tasks are not independent of each other. One person has to lift a heavy object and place it on a truck and the second person has to take the same object from the truck and place it on the floor. Both of these processes are dependent on each other. If one person stops lifting the object and the other person stops taking the object, then both get stuck without completing their tasks. Now let’s see the same example in a program. In a program, one process has to read a record from a database and place it in a certain table and the second process has to take the same record from the table and place it in another table. Both these tasks are not independent of each other. If one process stops reading a record from the database and the second process stops placing the record in another table, then both get stuck without completing their tasks. This is the deadlock condition.

Databases consist of logical, consistent, and organized data that can be accessed, managed, and updated easily. It is also known as an electronic database. Electronic databases are structured to provide the facility of creating, inserting, and updating data efficiently. They are typically stored on magnetic disks, tapes, or another sort of secondary storage medium. Objects in the database are called tables, and the records and fields are called records. An entity described by a database is described in terms of fields, which contain information about particular aspects or attributes of that entity. Database management systems are used to extract data from databases using queries.

Object-oriented programming (OOP) is a programming language design that focuses on modeling real-world objects as programming constructs like variables, functions, statements, etc. OOP is a way of breaking down a programming task by describing how different pieces of a program will interact with each other. It’s a concept that’s been around since the 1950s and is a common way for both beginner and advanced programmers to approach problems. Because OOP is such a foundational concept in computer programming, it’s important to understand what it is before moving on to other related concepts like object-based programming. OOP is built on a few basic concepts which allow programmers to create modular and reusable code. These concepts include Abstraction, Encapsulation, Inheritance, Polymorphism, and Hashes.

Unlike object-oriented programming, which defines the relationships between different types of code, object-based programming deals with the specific implementation of those codes. In other words, object-based programming is the process of writing code that’s been defined in advance as a particular type of object. This is a concept that’s been around since the invention of programming, but it’s been given a name in more recent years as OBP. OBP is an approach to software development that focuses on the creation of specific objects and how they work together. The goal of OBP is not just to create reusable code, but also to create a system of code that’s easy to maintain and extend in the future. The difference between OOP and OBP is found in their approach. While OOP is mostly concerned with creating reusable code, OBP is more concerned with creating a codebase that’s easy to maintain.

While OOP and OBP have a lot in common, there are also differences between the two concepts. Let’s take a look at what these differences are. Fundamentally, one of the biggest differences between OOP and OBP is that OOP is about creating reusable code, while OBP is about making code more sustainable for the future. OOP is a concept that’s been around since the invention of programming. OOP deals with the creation of code that’s modular and loosely coupled. OBP, on the other hand, is a more recent concept that’s been given its name in the last decade. OBP is all about creating sustainable code. Although OOP and OBP are different concepts with different focuses, they can both be used to solve the same problems.

The truth is, there’s no right answer when it comes to this question. Both OOP and OBP are valid approaches to solving problems, and each can be used in a variety of situations. However, the choice between OOP and OBP comes down to a few basic factors. First, you’ll want to consider the problem you’re trying to solve. Is there a specific type of problem you’re dealing with? If so, you can decide which concept is best for solving it. Next, you’ll want to consider your programming background. Are you a beginner programmer or an expert? If you’re a beginner, you’ll probably want to lean toward OOP, while an advanced programmer may lean toward OBP. Finally, you’ll want to consider your skill level. Are you comfortable with the way OOP works and how it’s implemented, or do you want to try something new?

Although data abstraction and data hiding are both important concepts in computer programming, they are different in a number of ways. Data abstraction is used to make certain implementation details irrelevant, while data hiding is used to make them inaccessible. Data abstraction is used to make code easier to maintain by keeping the code simple and only exposing what is necessary for it to work. Data hiding is used to make code easier to use by wrapping lower-level functionality in a way that is easier to work with. Data abstraction is used to make changes to code easier by not requiring users to know how something works, while data hiding is used to make changes easier by making it impossible for other parts of the code to access certain variables. Finally, data hiding is a great way to ensure that certain variables remain private and cannot be changed by other parts of the code, while data abstraction is a great way to make implementation details irrelevant so that users only need to see what they need to use the functionality.

• Practice More: You must have to practice the question on various coding platforms like - InterviewBit, LeetCode, Hackerrank,  etc. This will enhance your problem-solving skills.
• Learn Computer fundamentals: You must have an understanding of computer fundamental subjects like Operating systems, Database Management systems, Computer Networks, Data Structures, and object-oriented programming concepts. The interviewer expects answers from you from OOPs.
• Learn Java: If your primary language is java then it’s good. Otherwise, try to get some basic knowledge of java. Because HashedIn most of the work is based on Java.
• Analysis of Algorithms: Solve the Time complexity analysis of algorithms because the Interviewer highly expects this from you.
• Approach: In the coding interviews, always start with the brute force approach for solving the problems. Explain the approach quickly and then try to optimize it.

HasedIn follows the working environment like Deloitte because HashedIn is acquired by Deloitte. But both the firms work differently. HashedIn builds products for other enterprises and startups. HashedIn basically provides Product Engineering Services (PES).

HashedIn recruits the fresher candidates from On-Campus, Off-Campus, and also by Employee Referrals. There is another way to apply for HashedIn that is common for both freshers and experienced. That is from the HashedIn career portal.

In general, HashedIn Interview is not very difficult to clear. The interviewer's expectation for Software Engineers is better problem solvers. And if you are good at DSA-based problem solving, then you can crack it easily.

HashedIn offers 8.1 LPA CTC for freshers selected as Software Development Engineer - I. For an experienced person, selected as Software Development Engineer - II, HasedIn offers the CTC on average between 12 - 14 LPA.

HashedIn HR round (Fitment Round) is generally not an elimination round. HR basically wants to know more about you and the other offers in your hand with what CTC. And also HR checks your ability to interact with other persons.

The HashedIn work environment is great as described by the working professionals of HashedIn. Freshers have to graduate from HashedIn university during their Training. In this, freshers learn various new technologies and experience to write effective production-level code.

In each round of HashedIn Interview for Software engineers, The interview generally lasts for the duration of 1 - 1.5 hrs. And the HashedIn Fitment/HR round generally lasts for 0.5 hrs. After each round, you will get to know about your result whether you are selected for the round or not. The time HashedIn HR takes to respond is mostly between 24 - 72 hours. After the final result, you have to wait for at least 24 hours till you get the letter of intent for confirmation of your joining. Then within a week, your final offer letter will be released.

The basic eligibility criteria for Software Engineers in HashedIn by Deloitte are - 

• Qualification - B.E/B.Tech/M.E/M.Tech/MCA/M.Sc
• Percentage - 60% or 6.0 GPA above
• Backlog - No active backlogs