The document discusses data structures and algorithms. It defines data structures as a means of storing and organizing data, and algorithms as step-by-step processes for performing operations on data. The document also discusses abstract data types which define the operations that can be performed on a data structure independently of its specific implementation. Common data structures like stacks, queues, and lists are classified and their algorithms and applications explained.

1. Concept of Data Structure
 Data Structure:
A means of storing a collection of data.
OR
It is the specification of the elements of the
structure, the relationships between them, and
the operation that may be performed upon them.
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2. Computer Science & Data
Structure
Computer science is concern with study of
Methods for effectively using a computer to
solve problems.
or
In determining exactly the problem to be solved.
This process entails:
1. Gaining an understanding of the problem.
2. Translating vague descriptions, goals, and contradictory
requests, and often unstated desires, into a precisely
formulated conceptual Solution
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3. Computer Science & Data Structure
(Continue)
3. Implementing the solution with a computer
program. This solution typically consists of
two parts: algorithms and data structures:
Algorithm:
An algorithm is a concise specification of a method
for solving a problem.
Data Structure:
A data structure can be viewed as consisting of a set
of algorithms for performing operations on the data
it stores.
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4. Computer Science & Data Structure
(Continue)
Data Structure & Algorithm
Algorithms are part of what constitutes
a data structure. In constructing a solution to
a problem, a data structure must be chosen
that allows the data to be operated upon
easily in the manner required by the
algorithm.
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5.  12
 Data Structures vs. Algorithms
 •Choice of data structures
 �A primary design consideration at many times
 �Determines the algorithms to be used
 �“A well-designed data structure allows a variety of critical operations to be
performed using as few resources, both execution time and memory space, as
possible.”
 •The opposite direction sometimes
 �Certain algorithms require particular data structures
 �E.g., shortest path algorithm needs priority queues
– Data Structures
– How do we organize
– the data to be
– handled?
 Algorithms
 Describe the process
 by which data is
 handled.Duality
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6. 04/09/13 6

7. Abstract Data Types(ADT)
What is Abstraction?
To abstract is to ignore some details
of a thing in favor of others.
ADT & Data structure
An Abstract Data Type (ADT) is more a way of
looking at a data structure: focusing on what it
does and ignoring how it does its job.
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8. Abstract Data Types(ADT)
 Why need Abstraction?
Abstraction is important in problem solving because
it allows problem solvers to focus on essential details
while ignoring the inessential, thus simplifying the
problem and bringing to attention those aspects of the
problem involved in its solution. Once an abstract
data type is understood and documented, it serves as a
specification that programmers can use to guide their
choice of data representation and operation
implementation, and as a standard for ensuring
program correctness.
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9. Abstract Data Types(ADT)
Abstract data types:
Each data structure can be developed around the
concept of an abstract data type that defines both data
organization and data handling operations.
 A mathematical entity consisting of a set of values
and a collection of operations that manipulate them.
For example, the Integer abstract data type consists of
a carrier set containing the positive and negative
whole numbers and 0, and a collection of operations
manipulating these values, such as addition,
subtraction, multiplication, equality comparison, and
order comparison.
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10. Abstract Data Types(ADT)
 Abstract data types are important in computer science
because they provide a clear and precise way to
specify what data a program must manipulate, and
how the program must manipulate its data, without
regard to details about how data are represented or
how operations are implemented.
 ADT defines both data organization and data
handling operations.
 The study of data structure is organized around a
collection of abstract data types that includes lists,
trees, sets, graphs, and dictionaries.
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11. Abstract Data Types(ADT)
 ADT is not a part of a program ,since a
program written in a programming language
requires the definition of data structure, not
just the operations on the data structure.
 ADT is a useful tool for specifying the logical
properties of a data type.
 ADT is not concerned with the time/space
efficiency of data type.
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12. Specification of ADT
 ADT consist of 2 parts.
1. Value Definition.
2. Operator Definition.
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13. Specification of ADT
1. Value Definition.
It defines the collection of values for the
ADT and consist of two parts.
a) Definition Clause
b) Condition Clause
e.g. For ADT RATIONAL
Value definition: it states that RATIONAL value
consist of two integers ,the second of which does not equal
to zero.
The keyword abstract typedef introduce a value definition.
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14. Specification of ADT
 Condition Clause
 Used to specify any conditions on the newly defined
type.
 The keyword condition is used to specify any
conditions on the newly defined type.
For e.g. for RATIONAL ADT
Condition Clause: denominator may not be zero.
Note :The definition clause is required but the
condition clause may not be necessary for every
ADT.
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15. Specification of ADT
2. Operator Definition.
 It defines the operations that are to be
performed on a data set.
 Each operator is defined as an abstract
function with three parts:
 Header.
 Optional preconditions .
 Post conditions.
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16. Specification of ADT
2. Operator Definition.
For e.g. :Operator definition for RATIONAL ADT
includes the operations of creation ,addition ,
multiplication and equality.
 Now for multiplication operation the operator definition for
RATIONAL ADT is.
 Abstract RATIONAL mult(a,b)……/*header*/
RATIONAL a, b; ………. /*header*/
mul[0]=a[0]*b[0]; ………………….. /*postcond*/
mul[1]=a[1]*b[1];……………………/postcond*/
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17. Complete example of RATIONL
ADT
 /* value definition*/
 Abstract typedef< int , int >RATIONAL;
 Condition RATIONAL[1]!=0;
 /* Operator definition */
 Abstract equal(a,b)
 RATIONAL a ,b;
 Postcondition equal==(a[0]*b[1]==b[0]*b[1]);
 Like this we can declare the other operations.
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18. Classification of Data Structure
Linear and Non-Linear:
 In linear data structure, the data items are arranged
in linear sequence. e.g. Array, Stack, Queue, Linked
List.
 In non-linear data items are not in sequence. e.g.
Tree , graph ,Heap
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19. Classification of Data Structure
(Continue)
Homogenous and Non-homogenous.
 In homogenous data structure ,all the elements are
of same type e.g. Array.
 In non-homogenous data structure, the elements
may or may not be of same type e.g. Record.
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20. Classification of Data Structure
(Continue)
Primitive and Non-Primitive:
Primitive data structures constitute the
number and the characters which are built
programs. examples int. reals,pointer
data,charaters data, logical data.
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21. What is STACK or Pushdown
List
A ordered collection of items
into which new items may be
inserted and from which items
may be deleted at one end,
called the top of stack.
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22. Operations on STACK
 PUSH:When item is added in stack
 POP:When item is removed from
stack
 EMPTY:Stack containing no items.
 StackTop:Determine what the top
item on a stack without removing
it.this is a combination of push & pop
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23. Push & Pop in STACK
 PUSH Operation
 PUSH(STACK ,TOP,MAX,ITEM)
 STACK--Name given to Array.
 TOP----It contains the location of top
elements.
 MAX---It is maximum size of STACK.
 ITEM--->Value that is to be store in stack
if top = MAX then stackfull
top = top+1
stack(top) = item
Return
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24. Push & Pop in STACK
POP Operation
POP(STACK ,TOP,ITEM)
if top = 0 then Underflow
item = stack(top)
top = top-1
Return
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25. Applications of STACK
Expression Evaluation
Parenthesis Checker
Recursion
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26. Infix To Postfix Conversion of an
Arithmetic Expression
Operator preecedence
^ ---------Higher precedence
*,/--------Next Precedence.
+,_-------Least precedence.
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27. Infix To Postfix Conversion of an
Arithmetic Expression
 The rules to remember.
1. Parenthesize the expression
starting fro left to right.
2. During Parenthesizing the
expression the operands associated
with operator having higher
precedence are first parenthesized.
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28. Infix To Postfix Conversion of an
Arithmetic Expression
3. The sub-expression which has been
converted into postfix is to be
treated as single operand.
4. Once the expression is converted to
postfix form remove the
parenthesis.
Example: A+B*C
 A+[(B+C)+(D+E)*F]/G
 Answer: ABC+DE+F*+G/+
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29. Algorithm for Conversing Infix
Expression to Postfix Form
 Postfix(Q,P)
 Q—>Given Infix expression.
 P->Equivalent Postfix expression.
1. Push “(“ onto STACK and add “)” to the
end of Q.
2. Scan Q from left to right and repeat step
3 to 6 for each element of Q until the
STACK is Empty.
3. If an operand is encountered ,add it to P.
4. If a left parenthesis is encountered ,push
it onto STACK.
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30. Algorithm for Conversing Infix
Expression to Postfix Form
5. If an operator ® is encountered
then:
(a) ADD ® to STACK
[End of IF]
(b) Repeatedly pop from STACK and
add P each operator which has
same precedence as or higher
precedence than ®.
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31. Algorithm for Conversing Infix
Expression to Postfix Form
6. If a right parenthesis is encountered
then:
(a) Repeatedly pop from STACK and add to
P each operator until a left parenthesis is
encountered.
(b) Remove the left parenthesis. do not add
it of P]
[End of If]
[End of step 2 loop]
7. Exit.
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32. Converting Infix Expression to
Prefix Expression.
1. Reverse the input string.
2. Examine the next element in
the input.
3. If it is operand ,add it to the
output string.
4. If it is closing parenthesis ,
push it on STACK.
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33. Converting Infix Expression to
Prefix Expression.
5. If it is operator , then:
i) if STACK is empty, push –operation on
STACK.
ii) if the top of the stack is closing
parenthesis push operator on STACK.
iii) if it has same or higher priority then
the top of STACK, push operator on S.
iv) else pop the operator from the STACK
and add it to output string, repeat S.
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34. Converting Infix Expression to
Prefix Expression.
6. If it is a opening parenthesis , pop
operator from STACK and add them
to S until a closing parenthesis is
encountered. POP and discard the
closing parenthesis.
7. If there is more input go to step 2.
8. If there is more input , unstack the
remaining operators and add them.
9. Reverse the output string.
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35. Algorithm to Evaluate a Postfix
Expression
This algorithm finds the value of an
arithmetic expressions P written in
postfix notation. The algorithm uses a
STACK to hold operands , evaluate P.
1. Add right parenthesis “)” at the end
of P.[This acts as sentinel]
2. Scan P from left to right and repeat
step 3 & 4 for each element of P
until the sentinel “)” is
encountered.
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36. Algorithm to Evaluate a Postfix
Expression
3. If an operand is encountered , put it on
STACK.
4. If an operator ® is encountered , then:
A) Remove the two top elements of
STACK, Where A is the top element and
B is the next-to-top element.
B)Evaluate B ®A.
C)Place the result of (b) back o STACK.
[End of IF]
[End of step 2 loop]
6. Set value equal to the top element on
STACK.
7. Exit.
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37. Parenthesis Checker
 Read exp
 Create empty Stack
 For each character C in exp
 If(current character is left symbol) then
 Push the character C onto Stack.
 Elseif(current character C is a right
Symbol)then
 If(stack is empty)then
 Print:”Error:No matching open symbol”
 Exit
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38. Parenthesis Checker
 Else
 POP a symbol S from the Stack
 If (S doesnot correspond to C)then
 Print:”Error:Incorrect nesting of symbols”
 Exit
 Endif
 Endif
 Endif
 End for
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39. Parenthesis Checker
 If(stack is not empty)then
 Print:”Error:Missing closing
symbols(s)”
 Else
 Print:”Input expression is ok”
 Endif
 End
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40. Example of STACK
 Many computer algorithms work best with stacks --- stacks are
used for
 remembering partially completed tasks, and
 undoing (backtracking from) an action.
 An example of 1. is described in the next section, where sub
expressions of an arithmetic expression are remembered for later
computation. Another example is presented in the next lecture,
where we see how the Java Virtual Machine uses a stack to
remember all of a program's methods that have been called but
are not yet finished.
 An example of 2. is the ``undo'' button on most text editors,
which lets a person undo a typing error, or the ``back'' button on
a web browser, which lets a user backtrack to a previous web
page. Another example is a searching algorithm, which searches a
maze and keeps a history of its moves in a stack. If the algorithm
makes a false (bad) move, the move can be undone by retrieving
the previous position from the stack.
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41. STACK Implementation
 Static and dynamic data structures
 A stack can be stored in:
 • a static data structure
 OR
 • a dynamic data structure
 Static data structures
 These define collections of data which are fixed in size
when the program is compiled.
 An array is a static data structure.
 Dynamic data structures
 These define collections of data which are variable in size
and structure. They are
 created as the program executes, and grow and shrink to
accommodate the data being
 stored.
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42. TRANSPOSE OF SPARSE
MATRIX
 #include<stdio.h>
 #include<conio.h>
 // transpose for the sparse matrix
 void main()
 {
 //clrscr();
 int a[10][10],b[10][10];
 int m,n,p,q,t,col;
 int i,j;
 printf("enter the no of row and columns :n");
 scanf("%d %d",&m,&n);
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43. TRANSPOSE OF SPARSE
MATRIX
 // assigning the value of matrix
 for(i=1;i<=m;i++)
 {
 for(j=1;j<=n;j++)
 {
 printf("a[%d][%d]= ",i,j);
 scanf("%d",&a[i][j]);
 }
 }
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44. TRANSPOSE OF SPARSE
MATRIX
 printf("nn");
 //displaying the matrix
 printf("nnThe matrix is :nn");
 for(i=1;i<=m;i++)
 {
 for(j=1;j<=n;j++)
 {
 printf("%d",a[i][j]);
 }
 printf("n");
 }
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45. TRANSPOSE OF SPARSE
MATRIX
 t=0;
 printf("nnthe non zero value matrix are :nn");
 for(i=1;i<=m;i++)
 {
 for(j=1;j<=n;j++)
 {
// accepting only non zero value
 if(a[i][j]!=0)
 {
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46. TRANSPOSE OF SPARSE
MATRIX
 t=t+1;
 b[t][1]=i;
 b[t][2]=j;
 b[t][3]=a[i][j];
 } }
 }
 b[0][0]=m;
 b[0][1]=n;
 b[0][2]=t;
 // displaying the matrix of non-zero value
 printf("n");
 For(i=0;i<=t;i++)
 printf(“%d “,a[i][0],a[i][1],a[i][2]);
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47. TRANSPOSE OF SPARSE
MATRIX
 // transpose of the matrix
 printf("nnthe transpose of the matrix :n ");
 if(t>0)
 {
 for(i=1;i<=n;i++)
 {
 for(j=1;j<=t;j++)
 {
 if(b[j][2]==i)
 {
 a[q][1]=b[j][2]; a[q][2]=b[j][1];
 a[q][3]=b[j][3]; q=q+1;
 } }
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 } 47

48. TRANSPOSE OF SPARSE
MATRIX
 for(i=0;i<=t;i++)
 {
 printf("a[%d %d %d %dn",i,a[i][1],a[i][2],a[i][3]);
 }
 getch();
 }
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49. Tower of Hanoi(An application of
Recursion & Stack)
This problem is not specified in terms
of recursion but the recursive
technique can be used to produce a
logical solution .
 Problem Definition:
 There are three pegs A,B,& C exist,
any number of disks (n) of different
diameters are placed on peg A so
that a larger disk is always below a
smaller disk.
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50. Tower of Hanoi(An application of
Recursion & Stack)
 The object is to move n disks to peg
C ,using peg B as auxiliary.
 Only the top disk on any peg may be
moved to any other peg, and a
larger disk may never rest on a
smaller one.
 Note:in general it requires f(n)=2n-1
moves
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51. Recursive solution to tower of
Hanoi problem
 To move n disks from A to
C using B as auxiliary:
1. If n==1 move the single disk
from A to C and stop
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52. Recursive solution to tower of
Hanoi problem
2. Move the top n-1 disks from
A to B , using C as auxiliary.
3. Move the remaining disks
from A to C
4. Move the n-1 disks from B to
C,using A as auxiliary.
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53. Recursive Implementation of Tower
of Hanoi.
main()
{
int n;
scanf (“%d”, &n);
towers (n, ’A’, ’C’ ,’B’);
}
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54. towers(int n, char frompeg, char
topeg,char auxpeg)
{
If (n==1)
{
printf (“n %s %c %s %c”, “move
disk 1 from peg “,frompeg, “to
peg”, topeg);
return;
}04/09/13 54

55. towers(n-1,frompeg,auxpeg,topeg)
printf (“n %s %d %s %c %s %c”,
“move disk”, n, “from peg”, frompeg,
“to peg”, topeg);
tower(n-1,auxpeg,topeg,frompeg);
}
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