Summer2014 ASSIGNMENT
PROGRAM
- MCA (REVISED FALL 2012)
SEMESTER
- II
SUBJECT
CODE & NAME - MCA2020- ADVANCED DATA STRUCTURE
CREDIT – 4, BK ID - B1476, MAX. MARKS 60
Get fully solved assignment, plz drop a mail with your
sub code
computeroperator4@gmail.com
Charges rs 125/subject and rs 700/semester only.
our website is www.smuassignment.in
if urgent then call us on 08791490301, 08273413412
Q. No.1 Define data
structure? Explain different types of data structures. (2+8) 10 marks.
Answer: A data structure is a particular way of
storing and organizing data in a computer so that it can be used efficiently.
Different
kinds of data structures are suited to different kinds of applications, and
some are highly specialized to specific tasks. For example, B-trees are
particularly well-suited for implementation of databases, while compiler
implementations usually use hash tables to look up identifiers.
Q2. Describe the
following types of linked list. (5+5) 10 marks
a. Doubly linked list
Answer:
Doubly
linked list
In
some situation we need to traverse both forward and backward of a linked list.
The linked list with this property needs two link field one to point the next
node is called next link field and another to point the previous node is called
previous link field. The linked list containing this type of nodes is called
doubly linked list or two- way list. Here first nodes previous link field and
the
Q3. List the Advantages
and Disadvantages of Linear and linked representation of tree. [5+5] = 10
Answer:
Advantages of
linear representation of binary tree
1) Node can be accessed directly with the help of the index
through which we can improve the efficiency of execution time of an algorithm.
2
Q4. List and explain any
Five types of graph. [5*2]= 10
Answer:
Types of
Graphs
Depending upon
the vertices and edges and the weight associated to it, graphs can be
classified as:
Q5. Explain
1. Fixed block storage
allocation.
2. Variable block storage
allocation [5+5] = 10
Answer:
Fixed block
storage allocation
First block
storage allocation is the simplest case of dynamic storage allocation. This is
the straight forward method in which, all the blocks are of identical in size.
The user can decide the size of the block. The operating system keeps a pointer
called AVAIL. This
Q6. What is the use of
external Storage Devices? Explain any two external storage devices [4+3+3]=10
Answer:
An external storage device may be defined as device other than the
main memory on which information or data can be stored and from which the
information retrieved for processing. External storage devices are having
larger capacities and fewer expenses to store compared to main memory.
Get fully solved assignment, plz drop a mail with your
sub code
computeroperator4@gmail.com
Charges rs 125/subject and rs 700/semester only.
our website is www.smuassignment.in
if urgent then call us on 08791490301, 08273413412
No comments:
Post a Comment