SDE - 1

This can be solved both: recursively and iteratively.
The recursive approach is more intuitive. First reverse all the nodes after head. Then we need to set head to be the final node in the reversed list. We simply set its next node in the original list (head -> next) to point to it and sets its next to NULL. The recursive approach has a O(N) time complexity and auxiliary space complexity.

For solving the question is constant auxiliary space, iterative approach can be used. We maintain 3 pointers, current, next and previous, abbreviated as cur, n and prev respectively. All the events occur in a chain.
1. Assign prev=NULL, cur=head .
2. Next, repeat the below steps until no node is left to reverse:
1. Initialize n to be the node after cur. i.e. (n=cur->next)
2. Then make cur->next point to prev (next node pointer).
3. Then make prev now point to the cur node.
4. At last move cur also one node ahead to n.
The prev pointer will be the last non null node and hence the answer.

To reverse a number following steps should be performed:
Take the number’s modulo by 10
Multiply the reverse number by 10 and add modulo value into the reverse number.
Divide the number by 10.
Repeat above steps until number becomes zero.

Pseudocode :
reverse(n){

rev = 0; // reversed number
rem; // initialise a variable to store remainder

while(n>0){

rem = n%10;
rev = (rev*10) + rem;
n = n/10;
}

return rev;
}

This can be done by iterative swapping using two pointers. The first pointer points to the beginning of the string, whereas the second pointer points to the end. Both pointers keep swapping their elements and go towards each other. Essentially, the algorithm simulates the rotation of a string with respect to its midpoint.
Time Complexity : O(n)

One possible method is : 
1. Fill up the 5 litre bucket. 
2. Pour the 5 litre bucket into the 3 litre bucket until the 3 litre bucket is full.
3. Empty the 3 litre bucket.
4. Pour the remaining 2 litres from the 5 litre bucket into the 3 litre bucket.
5. Fill up the 5 litre bucket again.

Overloading allows different methods to have the same name, but different signatures where the signature can differ by the number of input parameters or type of input parameters or both. Overloading is related to compile-time (or static) polymorphism.

TELNET stands for TErminaL NETwork. It is a type of protocol that enables one computer to connect to local computer. It is a used as a standard TCP/IP protocol for virtual terminal service which is given by ISO. Computer which starts connection known as the local computer. Computer which is being connected to i.e. which accepts the connection known as remote computer.

1. SSH is more secure compared to Telnet
2. SSH encrypts the data while Telnet sends data in plain text
3. SSH uses a public key for authentication while Telnet does not use any authentication
4. SSH adds a bit more overhead to the bandwidth compared to Telnet
5. Telnet has been all but replaced by SSH in almost all uses.

The simple approach is to simply count the number of 0’s, 1’s, and 2’s. Then, place all 0’s at the beginning of the array followed by 1’s and 2's. The time complexity of this approach would be O(n) and space complexity O(1). 
However, the approach requires two traversals of the array. The question can also be solved in a single scan of the array by maintaining the correct order of 0’s, 1’s, and 2’s using variables. 
We divide the array into four groups using three pointers. Let us name these pointers as low, mid, and high.
1. a[0…low-1] only zeroes
2. a[low..mid-1] only ones
3. a[mid…high] unknown
4. a[high+1..n-1] only twos
Algorithm : 
1. Initialise three pointers low = 0, mid = 0 and high = n -1
2. Run a while loop until mid<=high : 
2.1 If (a[mid] ==0), then swap the element with the element low and increment low and mid (low++ and mid++).
2.2 If (a[mid] ==1), then increment mid (mid++).
2.3 If (a[mid] ==2), then swap it with an element in high range and decrement high (high--).

A hypervisor, also known as a virtual machine monitor or VMM, is software that creates and runs virtual machines (VMs). A hypervisor allows one host computer to support multiple guest VMs by virtually sharing its resources, such as memory and processing.

Cloud computing in very simple terms, is basically where a company uses someone elses computing services (usually over the internet) instead of having to run that software on their own computers.
Concrete examples are: in our very small business, we can't afford to have an IT department and people to manage our internal services. But we still need to use a variety of software, some simple examples are: accounting, file storage, shared calendars, shared contacts, etc. So we use a range of cloud computing services such as Dropbox, Google Calendar.

Stage 1: Planning and Requirement Analysis
Requirement analysis is the most important and fundamental stage in SDLC. It is performed by the senior members of the team with inputs from the customer, the sales department, market surveys and domain experts in the industry. This information is then used to plan the basic project approach and to conduct product feasibility study in the economical, operational and technical areas.

Stage 2: Defining Requirements
Once the requirement analysis is done the next step is to clearly define and document the product requirements and get them approved from the customer or the market analysts. This is done through an SRS (Software Requirement Specification) document which consists of all the product requirements to be designed and developed during the project life cycle.

Stage 3: Designing the Product Architecture
SRS is the reference for product architects to come out with the best architecture for the product to be developed. Based on the requirements specified in SRS, usually more than one design approach for the product architecture is proposed and documented in a DDS - Design Document Specification.
This DDS is reviewed by all the important stakeholders and based on various parameters as risk assessment, product robustness, design modularity, budget and time constraints, the best design approach is selected for the product.

Stage 4: Building or Developing the Product
In this stage of SDLC the actual development starts and the product is built. The programming code is generated as per DDS during this stage. If the design is performed in a detailed and organized manner, code generation can be accomplished without much hassle.
Developers must follow the coding guidelines defined by their organization and programming tools like compilers, interpreters, debuggers, etc. are used to generate the code. The programming language is chosen with respect to the type of software being developed.

Stage 5: Testing the Product
This stage is usually a subset of all the stages as in the modern SDLC models, the testing activities are mostly involved in all the stages of SDLC. However, this stage refers to the testing only stage of the product where product defects are reported, tracked, fixed and retested, until the product reaches the quality standards defined in the SRS.

Stage 6: Deployment in the Market and Maintenance
Once the product is tested and ready to be deployed it is released formally in the appropriate market. Sometimes product deployment happens in stages as per the business strategy of that organization. The product may first be released in a limited segment and tested in the real business environment (UAT- User acceptance testing).

A simple solution is to first find intersection of two arrays and store the intersection in a temporary array, then find the intersection of third array with the temporary array. 
Time complexity of this solution is O(n1 + n2 + n3) where n1, n2 and n3 are sizes of ar1[], ar2[] and ar3[] respectively.

The solution can be further optimised. Since all three arrays are sorted in strictly increasing order, it can be determined that if an array’s current integer is greater than another array’s current integer, then the another array’s current integer can’t be in the intersection.
Use three pointers in three arrays respectively to find the integers that appear in all three arrays. Initially, the three pointers point to the beginning of the three arrays respectively. Each time check the three integers that the three pointers point to. If a pointer points to an integer that is less than any of the other two integers, then move the pointer one step forward. Only if the three integers that three pointers point to are the same, the integer is in the intersection, and add the integer to the result list.

A volatile variable is a variable that is marked or cast with the keyword "volatile" so that it is established that the variable can be changed by some outside factor, such as the operating system or other software. A volatile variable is useful in multithreaded applications or in other situations where programmers must anticipate changes to the variable other than those that are common within the code module.

Multithreading is a Java feature that allows concurrent execution of two or more parts of a program for maximum utilization of CPU. Each part of such program is called a thread. So, threads are light-weight processes within a process.
Threads can be created by using two mechanisms : 
1. Extending the Thread class 
2. Implementing the Runnable Interface

Thread creation by extending the Thread class
We create a class that extends the java.lang.Thread class. This class overrides the run() method available in the Thread class. A thread begins its life inside run() method. We create an object of our new class and call start() method to start the execution of a thread. Start() invokes the run() method on the Thread object.

Thread creation by implementing the Runnable Interface
We create a new class which implements java.lang.Runnable interface and override run() method. Then we instantiate a Thread object and call start() method on this object.

Tip 1 : Be sure to do your homework on the organization and its culture before the interview.
Tip 2 : Employers want to understand how you use your time and energy to stay productive and efficient. Be sure to emphasize that you adhere to deadlines and take them seriously.
Tip 3 : Talk about a relevant incident that made you keen on the profession you are pursuing and follow up by discussing your education.