This document discusses data structures and algorithms. It begins by defining data structures as the logical organization of data and primitive data types like integers that hold single pieces of data. It then discusses static versus dynamic data structures and abstract data types. The document outlines the main steps in problem solving as defining the problem, designing algorithms, analyzing algorithms, implementing, testing, and maintaining solutions. It provides examples of space and time complexity analysis and discusses analyzing recursive algorithms through repeated substitution and telescoping methods.

1. 1
Data Structures

2. 2
The logical or mathematical model of a particular
organization of data is called a data structure
DATA STRUCTURES

3. 3
A primitive data type holds a single piece of data
–e.g. in Java: int, long, char, boolean etc.
–Legal operations on integers: + - * / ...
A data structure structures data!
–Usually more than one piece of data
–Should provide legal operations on the data
–The data might be joined together (e.g. in an array): a
collection
DATA STRUCTURES

4. 4
Static vs. Dynamic StructuresStatic vs. Dynamic Structures
A static data structure has a fixed size
This meaning is different than those associated with the static
modifier
Arrays are static; once you define the number of elements it can
hold, it doesn’t change
A dynamic data structure grows and shrinks as required by the
information it contains

5. 5
An Abstract Data Type (ADT) is a data type together
with the operations, whose properties are specified
independently of any particular implementation.
Abstract Data Type

6. 6
Abstract Data Type
In computing, we view data from three perspectives:
Application level
View of the data within a particular problem
Logical level
An abstract view of the data values (the domain) and the
set of operations to manipulate them
Implementation level
A specific representation of the structure to hold the data
items and the coding of the operations in a programming
language

7. 7
Problem Solving: Main Steps
1. Problem definition
2. Algorithm design / Algorithm specification
3. Algorithm analysis
4. Implementation
5. Testing
6. [Maintenance]

8. 8
Problem Definition
What is the task to be accomplished?
Calculate the average of the grades for a given student
What are the time / space / speed / performance requirements?

9. 9
. Algorithm Design / Specifications
Algorithm: Finite set of instructions that, if followed, accomplishes a
particular task.
Describe: in natural language / pseudo-code / diagrams / etc.
Criteria to follow:
Input: Zero or more quantities (externally produced)
Output: One or more quantities
Definiteness: Clarity, precision of each instruction
Finiteness: The algorithm has to stop after a finite (may be very
large) number of steps
Effectiveness: Each instruction has to be basic enough and
feasible

10. 10
Implementation, Testing, Maintenances
Implementation
Decide on the programming language to use
C, C++, Lisp, Java, Perl, Prolog, assembly, etc. , etc.
Write clean, well documented code
Test, test, test
Integrate feedback from users, fix bugs, ensure compatibility
across different versions  Maintenance

11. 11
Algorithm Analysis
Space complexity
How much space is required
Time complexity
How much time does it take to run the algorithm
Often, we deal with estimates!

12. 12
Space Complexity
Space complexity = The amount of memory required by an
algorithm to run to completion
[Core dumps = the most often encountered cause is “memory
leaks” – the amount of memory required larger than the
memory available on a given system]
Some algorithms may be more efficient if data completely loaded
into memory
Need to look also at system limitations
E.g. Classify 2GB of text in various categories [politics, tourism,
sport, natural disasters, etc.] – can I afford to load the entire
collection?

13. 13
Space Complexity (cont’d)
1. Fixed part: The size required to store certain data/variables, that
is independent of the size of the problem:
- e.g. name of the data collection
- same size for classifying 2GB or 1MB of texts
2. Variable part: Space needed by variables, whose size is
dependent on the size of the problem:
- e.g. actual text
- load 2GB of text VS. load 1MB of text

14. 14
Space Complexity (cont’d)
S(P) = c + S(instance characteristics)
c = constant
Example:
float sum (float* a, int n)
{
float s = 0;
for(int i = 0; i<n; i++) {
s+ = a[i];
}
return s;
}
Space? one word for n, one for a [passed by reference!], one
for i  constant space!

15. 15
Time Complexity
Often more important than space complexity
space available (for computer programs!) tends to be larger
and larger
time is still a problem for all of us
3-4GHz processors on the market
researchers estimate that the computation of various
transformations for 1 single DNA chain for one single protein
on 1 TerraHZ computer would take about 1 year to run to
completion
Algorithms running time is an important issue

16. 16
Running Time
Problem: prefix averages
Given an array X
Compute the array A such that A[i] is the average of elements
X[0] … X[i], for i=0..n-1
Sol 1
At each step i, compute the element X[i] by traversing the array
A and determining the sum of its elements, respectively the
average
Sol 2
At each step i update a sum of the elements in the array A
Compute the element X[i] as sum/I

17. 17
Running time
Input
1 ms
2 ms
3 ms
4 ms
5 ms
A B C D E F G
worst-case
best-case
}average-case?
Suppose the program includes an if-then statement that
may execute or not:  variable running time
Typically algorithms are measured by their worst case

18. 18
Experimental Approach
Write a program that implements the algorithm
Run the program with data sets of varying size.
Determine the actual running time using a system call to measure
time (e.g. system (date) );
Problems?

19. 19
Experimental Approach
It is necessary to implement and test the algorithm in order to
determine its running time.
Experiments can be done only on a limited set of inputs, and may
not be indicative of the running time for other inputs.
The same hardware and software should be used in order to
compare two algorithms. – condition very hard to achieve!

20. 20
Use a Theoretical Approach
Based on high-level description of the algorithms, rather than
language dependent implementations
Makes possible an evaluation of the algorithms that is
independent of the hardware and software environments

21. 21
Algorithm Description
How to describe algorithms independent of a programming
language
Pseudo-Code = a description of an algorithm that is
more structured than usual prose but
less formal than a programming language
(Or diagrams)
Example: find the maximum element of an array.
Algorithm arrayMax(A, n):
Input: An array A storing n integers.
Output: The maximum element in A.
currentMax ← A[0]
for i← 1 to n -1 do
if currentMax < A[i] then currentMax ← A[i]
return currentMax

22. 22
Properties of Big-Oh
Expressions: use standard mathematical symbols
use ← for assignment ( ? in C/C++)
use = for the equality relationship (? in C/C++)
Method Declarations: -Algorithm name(param1, param2)
Programming Constructs:
decision structures: if ... then ... [else ..]
while-loops while ... do
repeat-loops: repeat ... until ...
for-loop: for ... do
array indexing: A[i]
Methods
calls: object method(args)
returns: return value
Use comments
Instructions have to be basic enough and feasible!

23. 23
Asymptotic analysis - terminology
Special classes of algorithms:
logarithmic: O(log n)
linear: O(n)
quadratic: O(n2
)
polynomial: O(nk
), k ≥ 1
exponential: O(an
), n > 1
Polynomial vs. exponential ?
Logarithmic vs. polynomial ?

24. 24
Some Numbers
log n n n log n n
2
n
3
2
n
0 1 0 1 1 2
1 2 2 4 8 4
2 4 8 16 64 16
3 8 24 64 512 256
4 16 64 256 4096 65536
5 32 160 1024 32768 4294967296

25. 25
Relatives of Big-Oh
“Relatives” of the Big-Oh
Ω (f(n)): Big Omega – asymptotic lower bound
Θ (f(n)): Big Theta – asymptotic tight bound
Big-Omega – think of it as the inverse of O(n)
g(n) is Ω (f(n)) if f(n) is O(g(n))
Big-Theta – combine both Big-Oh and Big-Omega
f(n) is Θ (g(n)) if f(n) is O(g(n)) and g(n) is Ω (f(n))
Make the difference:
3n+3 is O(n) and is Θ (n)
3n+3 is O(n2
) but is not Θ (n2
)

26. 26
More “relatives”
Little-oh – f(n) is o(g(n)) if for any c>0 there is n0 such that f(n) <
c(g(n)) for n > n0.
Little-omega
Little-theta
2n+3 is o(n2
)
2n + 3 is o(n) ?

27. 27
Example
Remember the algorithm for computing prefix averages
compute an array A starting with an array X
every element A[i] is the average of all elements X[j] with j < i
Remember some pseudo-code … Solution 1
Algorithm prefixAverages1(X):
Input: An n-element array X of numbers.
Output: An n -element array A of numbers such that A[i] is the
average of elements X[0], ... , X[i].
Let A be an array of n numbers.
for i← 0 to n - 1 do
a ← 0
for j ← 0 to i do
a ← a + X[j]
A[i] ← a/(i+ 1)
return array A

28. 28
Example (cont’d)
Algorithm prefixAverages2(X):
Input: An n-element array X of numbers.
Output: An n -element array A of numbers such that A[i] is the
average of elements X[0], ... , X[i].
Let A be an array of n numbers.
s← 0
for i ← 0 to n do
s ← s + X[i]
A[i] ← s/(i+ 1)
return array A

29. 29
Back to the original question
Which solution would you choose?
O(n2
) vs. O(n)
Some math …
properties of logarithms:
logb(xy) = logbx + logby
logb (x/y) = logbx - logby
logbxa = alogbx
logba= logxa/logxb
–properties of exponentials:
a(b+c)
= ab
a c
abc
= (ab
)c
ab
/ac
= a(b-c)
b = a log
a
b
bc
= a c*log
a
b

30. 30
Important Series
Sum of squares:
Sum of exponents:
Geometric series:
Special case when A = 2
20
+ 21
+ 22
+ … + 2N
= 2N+1
- 1
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31. 31
Analyzing recursive algorithms
function foo (param A, param B) {
statement 1;
statement 2;
if (termination condition) {
return;
foo(A’, B’);
}

32. 32
Solving recursive equations by repeated substitution
T(n) = T(n/2) + c substitute for T(n/2)
= T(n/4) + c + c substitute for T(n/4)
= T(n/8) + c + c + c
= T(n/23
) + 3c in more compact form
= …
= T(n/2k
) + kc “inductive leap”
T(n) = T(n/2logn
) + clogn “choose k = logn”
= T(n/n) + clogn
= T(1) + clogn = b + clogn = θ(logn)

33. 33
Solving recursive equations by telescoping
T(n) = T(n/2) + c initial equation
T(n/2) = T(n/4) + c so this holds
T(n/4) = T(n/8) + c and this …
T(n/8) = T(n/16) + c and this …
…
T(4) = T(2) + c eventually …
T(2) = T(1) + c and this …
T(n) = T(1) + clogn sum equations, canceling
the terms appearing
on both sides
T(n) = θ(logn)

34. 34
RECURSION
Suppose P is a procedure containing either a CALL statement to
itself or a CALL statement back to original procedure P .Then P
is called a recursive procedure
Properties:
1. There must be certain criteria called basic criteria, for
which the procedure does not call itself.
2. Each time the procedure does call itself (directly or
indirectly), it must be closer to the base criteria.

35. 35
FACTORIAL WITHOUT RECURSION
FACTORIAL(FACT,N)
This procedure calculates N! and return the vale in the variable FACT
.
1. If N ==0,then :Set FACT:=1, and Return.
2. Set FACT:=1[Initialize FACT for loop]
3. Repeat for K:=1 to N
Set FACT:=K*FACT
[END of loop]
4. Return.

36. 36
FACTORIAL WITH RECURSION
FACTORIAL(FACT,N)
This procedure calculates N! and return the vale in the variable FACT .
1. If N ==0,then :Set FACT:=1, and Return.
2. Call FACTORIAL(FACT,N-1).
3. Set FACT:=N*FACT.
4. Return.

37. 37
FACTORIAL EXAMPLE USING RECURSION

38. 38
FACTORIAL EXAMPLE USING RECURSION

39. 39
FACTORIAL EXAMPLE USING RECURSION

40. 40
FACTORIAL EXAMPLE USING RECURSION

41. 41
FACTORIAL EXAMPLE USING RECURSION

42. 42
FACTORIAL EXAMPLE USING RECURSION

43. 43
FACTORIAL EXAMPLE USING RECURSION

44. 44
FACTORIAL EXAMPLE USING RECURSION

45. 45
FACTORIAL EXAMPLE USING RECURSION

46. 46
FACTORIAL EXAMPLE USING RECURSION

47. 47
FACTORIAL EXAMPLE USING RECURSION

48. 48
Stack
A stack is a list that has addition and deletion of items only from one
end.
It is like a stack of plates:
Plates can be added to the top of the stack.
Plates can be removed from the top of the stack.
This is an example of “Last in, First out”, (LIFO).
Adding an item is called “pushing” onto the stack.
Deleting an item is called “popping” off from the stack.

49. 49
STACK OPERATION (PUSH)
PUSH(STACK,TOP,MAXSTK,ITEM)
This procedure pushes an ITEM onto a stack.
1.[Stack already filled]
If TOP== MAXSTK, then: Print:OVERFLOW, and Return.
2. Set TOP:=TOP+1.[ Increases TOP by 1]
3. Set STACK[TOP]:=ITEM. [Inserting ITEM in new TOP position]
4. Return.

50. 50
STACK OPERATION (POP)
POP(STACK,TOP,ITEM)
This procedure deletes the top element of STACK and assigns it to the
variable ITEM .
1.[Stack has an item to be to removed]
If TOP== 0, then: Print:UNDERFLOW, and Return.
2. Set ITEM:=STACK[top].[ Assigns TOP element to ITEM ]
3. Set TOP:=TOP-1. [Decreases TOP by 1]
4. Return.

51. 51
STACK EXAMPLES
Implementing a stack :
MAXSTK=7.
Data
push 43
push 23
push 53
pop
pop
push 100

52. 52
STACK EXAMPLES
7
6
5
4
2
1
TOP=0
MAXSTK=7

53. 53
STACK EXAMPLES
7
6
5
4
2
1
TOP=1
MAXSTK=7
PUSH(43)
43

54. 54
STACK EXAMPLES
7
6
5
4
2
1
TOP=2
MAXSTK=7
PUSH(23)
43
23

55. 55
STACK EXAMPLES
7
6
5
4
2
1
TOP=3
MAXSTK=7
PUSH(53)
43
23
53

56. 56
STACK EXAMPLES
7
6
5
4
2
1
TOP=2
MAXSTK=7
POP( )
43
23

57. 57
STACK EXAMPLES
7
6
5
4
2
1
TOP=0
MAXSTK=7
POP( )
43

58. 58
STACK EXAMPLES
7
6
5
4
2
1
TOP=0
MAXSTK=7
PUSH(100)
43
100

59. 59
STACK EXAMPLES
1. INFIX to POSTFIX
2. POSTFIX expression solving
3. N-QUEEN’S problem

60. 60
INFIX to POSTFIX
Properties while transforming infix to postfix expression
1. besides operands and operators, arithmetic expression contains
left and right parentheses
2. We assume that the operators in q consist only of
1. Exponent
2. Multiplication
3. Division
4. Addition
5. Subtraction

61. 61
INFIX to POSTFIX
3. We have three levels of precedence
precedence operators
high right parentheses
exponent
multiplication, division
low addition, subtraction

62. 62
INFIX to POSTFIX
POLISH(Q, P)
Suppose Q is an arithmetic expression written in infix notation.
This algorithm finds the equivalent postfix expression P.
1. Push “(“ into 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.
5. If an operator is encountered then:
(a) repeatedly pop from STACK and to P each operator
(on the top of STACK) which has the same precedence as or
higher precedence this operator
(b) Add operator to STACK

63. 63
INFIX to POSTFIX
6. If a right parenthesis is encountered then:
(a) Repeatedly pop from STACK and add to P each
operator( on the top of STACK) until a left parenthesis is
encountered
(b) Remove the left parenthesis .[Do not add the left
parenthesis to P]
7. Exit

64. 64
INFIX to POSTFIX
INFIX Expression: 3 + 2 * 4
POSTFIX Expression:
7
6
5
4
2
1

65. 65
INFIX to POSTFIX
INFIX Expression: + 2 * 4
POSTFIX Expression: 3
7
6
5
4
2
1

66. 66
INFIX to POSTFIX
INFIX Expression: 2 * 4
POSTFIX Expression: 3
7
6
5
4
2
1 +

67. 67
INFIX to POSTFIX
INFIX Expression: * 4
POSTFIX Expression: 3 2
7
6
5
4
2
1 +

68. 68
INFIX to POSTFIX
INFIX Expression: 4
POSTFIX Expression: 3 2
7
6
5
4
2
1 +
*

69. 69
INFIX to POSTFIX
INFIX Expression:
POSTFIX Expression: 3 2 4
7
6
5
4
2
1 +
*

70. 70
INFIX to POSTFIX
INFIX Expression:
POSTFIX Expression: 3 2 4 *
7
6
5
4
2
1 +

71. 71
INFIX to POSTFIX
INFIX Expression:
POSTFIX Expression: 3 2 4 * +
7
6
5
4
2
1

72. 72
POSTFIX EXPRESSION SOLVING
This algorithm finds the VALUE of an arithmetic expression P
written in postfix notation
1. Add a right parenthesis “)” at the end of P
2. Scan P from left to right and repeat step 3 and 4 for each
element of P until the sentinel “)” is encountered.
3. If an operand is encountered, put it on STACK.
4. If an operator is uncounted then:
a. Remove the two elements of Stack, where A is the top
element and B is the next top element
b. Evaluate B operator A
c. Pace the result of (b) back on STACK

73. 73
POSTFIX EXPRESSION SOLVING
5. Set VALUE equal to the top element on STACK
6. Exit

74. 74
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 3 2 4 * +
Push the numbers until operator comes
7
6
5
4
2
1

75. 75
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 2 4 * +
7
6
5
4
2
1 3

76. 76
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 4 * +
7
6
5
4
2
1 3
2

77. 77
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: * +
Here we pop two time and perform
multiplication on elements (4*2) and
push the Result in to the stack
7
6
5
4
2
1 3
4
2

78. 78
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: +
7
6
5
4
2
1 3
8

79. 79
POSTFIX EXPRESSION SOLVING
POSTFIX Expression: 3 2 4 * +
7
6
5
4
2
1 11

80. 80
The N-Queens Problem
Can the queens be placed on the
board so that no two queens are
attacking each other in chess
board

81. 81
The N-Queens Problem
Two queens are not allowed in the same rowTwo queens are not allowed in the same row

82. 82
The N-Queens Problem
Two queens are not allowed in the same row, or in the same column...Two queens are not allowed in the same row, or in the same column...

83. 83
The N-Queens Problem
Two queens are not allowed inTwo queens are not allowed in
the same row, or in the samethe same row, or in the same
column, or along the samecolumn, or along the same
diagonal.diagonal.r along the same
diagonal.

84. 84
The N-Queens Problem
The number of queens,
and the size of the board
can vary.
The number of queens,
and the size of the board
can vary.
N Queens
N columns

85. 85
The 3-Queens Problem
The program uses aThe program uses a
stack to keep track ofstack to keep track of
where each queen iswhere each queen is
placed.placed.

86. 86
The 3-Queens Problem
Each time the program
decides to place a
queen on the board,
the position of the new
queen is stored in a
record which is placed
in the stack.
ROW 1, COL 1

87. 87
The 3-Queens Problem
We also have an integer
variable to keep track of
how many rows have been
filled so far.
ROW 1, COL 1
1
filled

88. 88
The 3-Queens Problem
Each time we try to place
a new queen in the next
row, we start by placing
the queen in the first
column
ROW 1, COL 1
1
filled
ROW 2, COL 1

89. 89
The 3-Queens Problem
if there is a conflict with
another queen, then we
shift the new queen to
the next column.
ROW 1, COL 1
1
filled
ROW 2, COL 2

90. 90
The 3-Queens Problem
ROW 1, COL 1
1
filled
ROW 2, COL 3
When there are noWhen there are no
conflicts, we stop andconflicts, we stop and
add one to the value ofadd one to the value of
filled.filled.

91. 91
The 3-Queens Problem
When there are no
conflicts, we stop and
add one to the value of
filled.
ROW 1, COL 1
2
filled
ROW 2, COL 3

92. 92
The 3-Queens Problem
Let's look at the third row.
The first position we try has
a conflict
ROW 1, COL 1
2
filled
ROW 2, COL 3
ROW 3, COL 1

93. 93
The 3-Queens Problem
so we shift to column 2.
But another conflict
arises
ROW 1, COL 1
2
filled
ROW 2, COL 3
ROW 3, COL 2

94. 94
The 3-Queens Problem
and we shift to the third
column.
Yet another conflict arises
ROW 1, COL 1
2
filled
ROW 2, COL 3
ROW 3, COL 3

95. 95
The 3-Queens Problem
and we shift to column 4.
There's still a conflict in
column 4, so we try to
shift rightward again
ROW 1, COL 1
2
filled
ROW 2, COL 3
ROW 3, COL 4

96. 96
The 3-Queens Problem
but there's
nowhere else to
go.
ROW 1, COL 1
2
filled
ROW 2, COL 3
ROW 3, COL 4

97. 97
The 3-Queens Problem
When we run out of
room in a row:
 pop the stack,
 reduce filled by 1
 and continue
working on the previous
row.
ROW 1, COL 1
1
filled
ROW 2, COL 3

98. 98
The 3-Queens Problem
Now we continue working
on row 2, shifting the
queen to the right.
ROW 1, COL 1
1
filled
ROW 2, COL 4

99. 99
The 3-Queens Problem
This position has no
conflicts, so we can
increase filled by 1, and
move to row 3.
ROW 1, COL 1
2
filled
ROW 2, COL 4

100. 100
The 3-Queens Problem
In row 3, we start again
at the first column.
ROW 1, COL 1
2
filled
ROW 2, COL 4
ROW 3, COL 1

101. 101
The 3-Queens Problem
In row 3, we start again
at the first column.
ROW 1, COL 1
3
filled
ROW 2, COL 4
ROW 3, COL 2

102. 102
QUEUES
A queue is a data structure that maintains a “first-in first-out”
(FIFO) ordering.
In contrast, a stack maintains a “last-in first-out” (LIFO) ordering.
A queue adds new elements at the end. An element can only be
removed at the front.
This is an abstraction of the “first-come first-served” practice.

103. 103
QUEUE OPERATIONS
A queue has two operations:
QINSERT
QDELETE
An enqueue operation adds new elements at the end of the queue
or its tail. This is similar to the stack operation push; only that push
now is done at the end of the array instead of at the front (or top).
A dequeue operation removes an element from the front of the
array or its head.

104. 104
QUEUE INSERTION
QINSERT(rear, item)
1. IF rear == MAX_QUEUE_SIZE_1 then
Print queue_full
Return;
2. INFO [++rear] = item;

105. 105
QUEUE DELETION
QDELETE(front, rear)
1. IF front == rear then
Return queue_empty( );
Return queue [++ front];

106. 106
QUEUE
OFFSET
DATA
0 1 2 3
0
rear
0
OFFSET
DATA
0 1 2 3
A
front
1
rear
1
OFFSET
DATA
0 1 2 3
A B
front
1
rear
2
OFFSET
DATA
0 1 2 3
B
front
2
rear
2
front
Insert A
Insert B
delete

107. 107
CIRCULAR QUEUE
When a new item is inserted at the rear, the to rear moves
upwards.
Similarly, when an item is deleted from the queue the front arrow
moves downwards.
After a few insert and delete operations the rear might reach the
end of the queue and no more items can be inserted although
the items from the front of the queue have been deleted and
there is space in the queue.

108. 108
CIRCULAR QUEUE
To solve this problem, queues implement wrapping around. Such
queues are called Circular Queues.
Both the front and the rear wrap around to the beginning of the
array when they reached the MAX size of the queue.
It is also called as “Ring buffer”.

109. 109
CIRCULAR QUEUE INSERTION
QINSERT (QUEUE, N, FRONT, ITEM)
This procedure insert an element ITEM into a queue.
1. [Queue already filled?]
IF ( FRONT==1 and REAR==N ) or FRONT ==REAR + 1,then:
write: overflow, and Return
2.[Find new value of REAR]
IF FRONT==NULL then [Queue initially empty.]
Set FRONT=1 and REAR=1
ELSE IF REAR ==N then
Set REAR=1
ELSE
set REAR=REAR+1
[End of if structure]

110. 110
CIRCULAR QUEUE INSERTION
3. Set QUEUE[REAR]=ITEM.[This inserts new element]
4. Return.

111. 111
CIRCULAR QUEUE DELETION
QDELETE(QUEE, N, FRONT, REAR, ITEM)
This procedure deletes an element from a queue and assigns it to the
variable ITEM
1.[Queue already empty]
if FRONT=NULL then write UNDERFLOW, and Return
2. Set ITEM=QUEUE[FRONT]
3. [Find new value of FRONT]
If FRONT =REAR then [Queue has only one element to start]
Set FRONT=NULL and REAR=NULL

112. 112
CIRCULAR QUEUE DELETION
Else if FRONT ==N then
Set FRONT=1
Else
Set FRONT=FRONT +1
[End of if statement]
4.Return

113. 113
CIRCULAR QUEUE DELETION
EMPTY QUEUE
[2] [3] [2] [3]
[1] [4] [1] [4]
[0] [5] [0] [5]
front = 0 front = 0
rear = 0 rear = 3
J2
J1
J3

114. 114
CIRCULAR QUEUE DELETION
[2] [3] [2] [3]
[1] [4][1] [4]
[0] [5] [0] [5]
front =0
rear = 5
front =4
rear =3
J2 J3
J1 J4
J5 J6 J5
J7
J8 J9

115. 115
PRIORITY QUEUE
A priority queue is a collection of elements such that each
element has been assigned a priority and such that the order
in which elements are deleted and processed comes from the
following rules
1. An element of a higher priority is processed before any
elements of lower priority
2. Two elements with the same priority are processed according
to the order in which they were added to the queue

116. 116
ARRAY REPRESENTATION PRIORITY QUEUE
Use a separate queue for each level of priority
Each such queue will appear in its own circular array and must
have its own pair of pointers FRONT and REAR
In fact, if each queue is allocated the same amount of space in
two-dimensional array QUEUE can be used instead of linear
array

117. 117
DELETION ON PRIORITY QUEUE
Deletion
This algorithm deletes and processes the first element in a
priority queue maintained by a two-dimensional array QUEUE
1.[Find the first nonempty queue]
Find the smallest k that FRONT!=NULL
2.Delete and process the front element in row K of QUEUE

118. 118
INSERTION ON PRIORITY QUEUE
Insertion
This algorithm adds an ITEM with priority number M to a priority
queue maintained by a two-dimensional array QUEUE
1. Insert ITEM as the rear element in row M of QUEUE
2. Exit

119. 119
LINKED LIST
Linked list
Linear collection of self-referential class objects, called
nodes
Connected by pointer links
Accessed via a pointer to the first node of the list
Subsequent nodes are accessed via the link-pointer member
of the current node
Link pointer in the last node is set to null to mark the list’s
end
Use a linked list instead of an array when
You have an unpredictable number of data elements
Your list needs to be sorted quickly

120. 120
TYPES OF LINKED LIST
Types of linked lists:
Singly linked list
Begins with a pointer to the first node
Terminates with a null pointer
Only traversed in one direction
Circular, singly linked
Pointer in the last node points back to the first node
Doubly linked list
Two “start pointers” – first element and last element
Each node has a forward pointer and a backward pointer
Allows traversals both forwards and backwards
Circular, doubly linked list
Forward pointer of the last node points to the first node
and backward pointer of the first node points to the last
node

121. 121
linked list
Start Node
A placeholder node at the beginning of the list, used to simplify
list processing. It doesn’t hold any data
Tail Node
A placeholder node at the end of the list, used to simplify list
processing

122. 122
Singly linked list
Single Linked List
Consists of data elements and reference to the
next Node in the linked list
First node is accessed by reference
Last node is set to NULL
A B C
Start Tail

123. 123
SINGLE LINKED LIST INSERTION
INSFIRST(INFO, LINK, START, AVAIL, ITEM)
This algorithm inserts ITEM as the first node in the list
1. [OVERFLOW?] If AVAIL=NULL then :Write OVERFLOW
and Exit
2. [Remove first node from AVIL list]
Set NEW =AVAIL and AVAIL=LINK[AVAIL]
3. Set INFO[NEW]=ITEM [Copies new data into new node]
4. Set LINK[NEW]=START [new node now points to original
first node]
5. Set START=NEW [Changes START so it points to the
node]
6. Exit

124. 124
SINGLE LINKED LIST INSERTION
s Xcda
current
1
start

125. 125
SINGLE LINKED LIST INSERTION
s Xcda
current
1
start

126. 126
SINGLE LINKED LIST INSERTION
s Xcda
current
1
start

127. 127
SINGLE LINKED LIST INSERTION
s Xcda
current
1
start

128. 128
SINGLE LINKED LIST INSERTION
s Xcda
current
1
start

129. 129
SINGLE LINKED LIST INSERTION
INSLOC(INFO, LINK,START, AVAIL, LOC, ITEM)
This algorithm inserts ITEM so that ITEM follows the node with
location LOC or inserts ITEM as the first node when LOC=NULL
1. [OVERFLOW ?] If AVAIL==NULL then write OVERFLOW and
EXIT
2. [Remove first node from AVAIL list]
3. Set INFO[NEW]=ITEM[copies new data into new node]
4. If LOC==NULL then :[insert as first node]
Set LINK[NEW]=Start and START=NEW
else [insert after node with location LOC]
Set LINK[NEW]=LINK[LOC] and LINK[LOC]=NEW
end of if structure
5. exit

130. 130
SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current

131. 131
SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current

132. 132
SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current

133. 133
SINGLE LINKED LIST INSERTION
9
6 X532start
temp
current

134. 134
SINGLE LINKED LIST DELETION
DELETE(INFO, LINK,START, AVAIL, ITEM)
This algorithm deletes from a linked list the first node N which
contains the given ITEM of information
1. [Use procedure FIND given after this algorithm]
FIND(INFO, LINK, START, ITEM, LOC, LOCP)
2. If LOC==NULL then:
Write ITEM not in list and exit
3. If LOCP==NUL then
Set START=LINK[START]
else :
Set LINK[LOCP]=LINK[LOC]
3. {Return deleted node to the AVAIL list]
Set LINK[LOC]=AVAIL and AVAIL=LOC
4. Exit

135. 135
SINGLE LINKED LIST DELETION
FINDB(INFO, START, ITEM, LOC, LOCP)
This procedure finds the location LOC of first node N which
contains ITEM and the location LOVP of the node preceding N. if
ITEM does not appear in the list then the procedure sets
LOC=NULL and if ITEM appears in the first node then it sets
LOCP=NULL
1. [list empty?] if START==NULL then
Set LOC =NULL and LOCP==NULL and return
2. [ITEM in first node ] if INFO[START]==ITEM then
Set LOC=START and LOCP=NULL and return
3. Set SAVE=START and PRT=LINK[START]
4. Repeat steps 5 and 6 while PTR!=NULL
5. If INFO[PTR]==ITEM then
Set LOC=PTR and LOCP=NULL

136. 136
SINGLE LINKED LIST DELETION
6. Set SAVE=PTR and PTR=LINK[PTR]
7. Set LOC=NULL
8. Return

137. 137
SINGLE LINKED LIST DELETION
6 X532start
LOCP
LOC

138. 138
SINGLE LINKED LIST DELETION
6 X532start
LOCP
LOC

139. 139
SINGLE LINKED LIST DELETION
6 X532start
LOCP
LOC

140. 140
Double Linked Lists
Consists of nodes with two linked references one points to the
previous and other to the next node
Maximises the needs of list traversals
Compared to single list inserting and deleting nodes is a bit
slower as both the links had to be updated
It requires the extra storage space for the second list
A B C
start
Tail

141. 141
Insertion Double Linked Lists
INSERTWL(INFO,FORW, BACK, STACK,AVAIL,LOCA,LOCB, ITEM)
1. [OVERFLOW?] If AVAIL==NULL then write OVER FLOW and exit
2. [Remove node from AVAIL list and copy new data into node]
Set NEW=AVIAL, AVIAL=FORW[AVAIL] INFO[NEW]=ITEM
3. [Insert node into list]
Set FORW[LOCA]=NEW, FORW[NEW]=LOCB
BACK[LOCB]=NEW BACK[NEW]=LOCA
4. Exit

142. 142
Insertion Double Linked Lists
3 4 7
start
Tail
Node A Node B
LOC A LOC B
7NEW

143. 143
Insertion Double Linked Lists
3 4 7
start
Tail
Node A Node B
LOC A LOC B
7NEW

144. 144
Insertion Double Linked Lists
3 4 7
start
Tail
Node A Node B
LOC A LOC B
7NEW

145. 145
Insertion Double Linked Lists
3 4 7
start
Tail
Node A Node B
LOC A LOC B
7NEW

146. 146
Insertion Double Linked Lists
3 4 7
start
Tail
Node A Node B
LOC A LOC B
7NEW

147. 147
Deletion Double Linked Lists
DELTWL(INFO, FORW, BACK, START, AVAIL, LOC)
1. [Delete node]
Set FORE[BACK[LOC]]=FORW[LOC]
Set BACK[FORW[LOC]]=BACK[LOC]
2. [Return node to AVAIL list]
Set FORW[LOC]=AVAIL and AVAIL=LOC
3. Exit

148. 148
Deletion Double Linked Lists
A B C
start
Tail
LOC

149. 149
Deletion Double Linked Lists
A B C
start
Tail
LOC

150. 150
Deletion Double Linked Lists
A B C
start
Tail

151. 151
Deletion Double Linked Lists
A B C
start
Tail