Software Engineer Trainee

Tip 1: Practice a lot of the questions on the Indiabix website. 

Tip 2: If you're short on time and can't practice answering a lot of questions, at least view the formula and memorize it (e.g. simple interest, compound interest). 

Tip 3: Be quick in solving questions, as I've heard that many applicants are turned down by recruiters here. Hence, achieving a high score in this area is crucial.

Initialize the first as 0(i.e, index of arr[0] element
Start traversing the array from array[1],
If the current element in the array say arr[i] is greater than the first. Then update first and second as, second = first and first = arr[i]
If the current element is in between the first and second, then update the second to store the value of a current variable as second = arr[i]
Return the value stored in the second.

Follow the given steps to solve the problem:

1. Create a function permute() with parameters: input string, starting index of the string, ending index of the string.
2. Call this function with values: input string, 0, size of string – 1.
3. In this function, if the value of L and R is the same, then print the same string.
4. Else, run a for loop from L to R and swap the current element in the loop with the inputString[L].
5. Then, call this same function again by increasing the value of L by 1.
6. After that, swap the previously swapped values to initiate backtracking.
 

I was asked about the question. But they don't want me to solve it properly. they wanted only to tell the approach.
I told them about the brute force approach and I was asked to find the time complexity of the approach
I suggested to them I could solve this issue in less time. 

But interviewer posted a new question in Zoom chat

I was asked to explain the whole flow of the binary search algorithm: 

how it works, how it is helpful, and why we use it. 

Additionally, I was asked about its time complexity and why it is lower compared to linear search and many other questions.

A simple method is to create a Hash Table. Calculate the hash value of each word in such a way that all anagrams have the same hash value. Populate the Hash Table with these hash values. Finally, print those words together with the same hash values. A simple hashing mechanism can be modulo the sum of all characters. With modulo sum, two non-anagram words may have the same hash value. This can be handled by matching individual characters.

Tip 1: Go through your college semester notes on these subjects and you are good to go.
 

Tip 1: Answer all questions confidently.
Tip 2: Go through top 50 questions asked in an interview about each topic from Google.
 

They convert primitive data types into objects. Objects are needed if we wish to modify the arguments passed into a method (because primitive types are passed by value).

Stack Memory
The stack memory is a physical space (in RAM) allocated to each thread at run time. It is created when a thread is created. Memory management in the stack follows the LIFO (Last-In-First-Out) order because it is accessible globally. It stores the variables, references to objects, and partial results. Memory allocated to stack lives until the function returns. If there is no space for creating the new objects, it throws the java.lang.StackOverFlowError. The scope of the elements is limited to their threads. The JVM creates a separate stack for each thread.

Heap Memory
It is created when the JVM starts up and is used by the application as long as the application runs. It stores objects and JRE classes. Whenever we create objects it occupies space in the heap memory while the reference of that object is created in the stack. It does not follow any order like the stack. It dynamically handles the memory blocks. It means we do need not to handle the memory manually. For managing the memory automatically, Java provides a garbage collector that deletes the objects that are no longer being used. Memory allocated to heap lives until any one event, either program terminated or memory does not occur. The elements are globally accessible in the application. It is a common memory space shared with all the threads. If the heap space is full, it throws the java.lang.OutOfMemoryError.