Test: Programming & Data Structures - 1 - Computer Science Engineering (CSE) MCQ

The basic character set of FORTRAN contains -

1. Lower case alphabets: a to z
2. Upper case alphabets: A to Z
3. Digits: 0 to 9
4. Underscore: _

.exrc: It reads the file and applies the settings in the file being opened.

.profile: It contains your individual profile that overrides the variables set in the /etc/profile file, hence, setting the environment for the user.

.lastlogin: There is no such file.

.mbox: It contains the family of related file formats used for holding collections of email messages.

• The C language is also known as the mother of all programming languages.
• C is a general-purpose programming language that is used for creating a variety of applications.
• C language was originally developed for writing operating systems. Unix Kernel and all of its supporting tools and libraries are written in C language.
• The C language is used for the following operations :
  • Operating systems
  • Development of new languages
  • Computation platforms
  • Embedded systems
  • Graphics and Games
• C++ and Java are the high-level languages and COBOL is a compiled English-like computer programming language.

Additional Information

• Computer Language - A programming language is a vocabulary and set of grammatical rules for instructing a computer or computing device to perform specific tasks.
• Operating System - is system software that manages computer hardware and software resources and provides common services for other computer programs.

CONCEPT:

Structure in C is a user-defined data type that is used to store the collection of different data types.

The total size of the structure is the sum of the size of every data member.

Union is a user-defined data type that is used to store different data types in the same memory location.

The total size of the union is the size of the largest data member.
 

Given the size of short, float and long is 2 bytes, 4 bytes, and 8 bytes, respectively.

Therefore,

Size of Structure → size of ( short s[5] ) + size of Union  

Here, Size of Union = 8 bytes   { largest data member is long }.

Size of short s[5] → 2×5 → 10 bytes

Size of Structure → 10+8 →18 bytes.

Concept

The javac command in Java compiles a program from a command prompt.

It reads a Java source program from a text file and creates a compiled Java class file.

Syntax

javac filename [options]

For example, to compile a program named Abc.java, use this command:

javac Abc.java

Both Merge sort and Insertion sort can be used for linked lists.

The slow random-access performance of a linked list makes other algorithms (such as quicksort) perform poorly, and others (such as heapsort) completely impossible.

Since worst case time complexity of Merge Sort is O(nLogn) and Insertion sort is O(n^2), merge sort is preferred.

See following for implementation of merge sort using Linked List.

To keep the Last In First Out order, a stack can be implemented using linked list in two ways:

a) In push operation, if new nodes are inserted at the beginning of linked list, then in pop operation, nodes must be removed from beginning.

b) In push operation, if new nodes are inserted at the end of linked list, then in pop operation, nodes must be removed from end.

Suppose we start filling the queue.

Let the maxQueueSize ( Capacity of the Queue) is 4. So the size of the array which is used to implement this circular queue is 5, which is n.

In the beginning when the queue is empty, FRONT and REAR point to 0 index in the array.

REAR represents insertion at the REAR index.
FRONT represents deletion from the FRONT index.

enqueue("a"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 1)
enqueue("b"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 2)
enqueue("c"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 3)
enqueue("d"); REAR = (REAR+1)%5; ( FRONT = 0, REAR = 4)

Now the queue size is 4 which is equal to the maxQueueSize. 
Hence overflow condition is reached.

Now, we can check for the conditions.

When Queue Full :

(REAR+1)%n = (4+1)%5 = 0

FRONT is also 0.

Hence ( REAR + 1 ) %n is equal to FRONT.

When Queue Empty :

REAR was equal to FRONT when empty ( because in the starting 
before filling the queue FRONT = REAR = 0 )

Hence Option A is correct.

In-order traversal of a BST gives elements in increasing order. So answer c is correct without any doubt.