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Data Structures
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Prof. Dr. Wajid Aziz
kh.wajid@ajku.edu.pk
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Motivation
• You want to store 4 numbers in a program
o No problem. You define four float variables:
float num1, num2, num3, num4;
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o Easy enough, right?
o But what if you want to store 2000 numbers?
• Are you really going to make 2000 separate variables?
float num1, num2,..., num1998, num1999,num2000;
• That would be CRAZY!
• So, what is the solution?
o A data structure! Specifically, an array!
• An array is one of the most common data structures.
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Basic Concepts
• Array name (data)
• Index/subscript (0...9)
• The slots are numbered sequentially
starting at zero (Java, C++)
• If there are N slots in an array, the
index will be 0 through N-1
• Array length = N = 10
• Array size = N x Size of an element = 40
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Using Arrays
• Array_name[index]
• For example, in C++
ocout<<data[4];
• will display 0
odata[3] = 99;
• Will replace -3 with 99
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Using Arrays
• data[ -1 ]
o illegal
• data[ 10 ]
 What will be the output of?
 data[5] + 10
 data[3] = data[3] + 10
o illegal (10 > upper bound)
• data[ 1.5 ]
o illegal
• data[ 0 ]
o OK
• data[ 9 ]
o OK
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Array As a Data Structure
• In general, the length of an arrayA be obtained from
index set formula
Length=UB-LB+1
• Where LB is lower bound (smallest index), UB is upper
bound (largest index) and The simplest type of data
structure is linear array.
• Example: An automobile company uses an array AUTO
to record number of automobiles sold each years from
1932 to 1984.
o Rather beginning index set with 0, it more convenient to
begin from 1932, then
• LB=1932, UB=1984, hence length of array will be
Length = UB-LB+1 = 1984-1932+1 = 53
Prof. Dr. Wajid Aziz
Array As a Data Structure
• The simplest type of data structure is linear array.
o It is a collection of multiple values of same type
o Examples:
o An array of student grades
o An array of student names
o An array of objects (OOP perspective!)
• If we choose name A for the array, then array
elements are denoted by
A[1], A[2], A[3], …, A[N]
• The K in A[K] is called subscript and A[K], is called
subscripted variable.
• The number N called length or size of array.
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Representation of Arrays in Memory
• Let LA be a linear array in the memory of the 1000
computer, then
LOC(LA(K))= Address of the element LA(K) of array LA
1001
1002
• The computer does not need to keep track of
1003
the address of every element of LA
o Needs to keep track the address of first 1004
element of LA, called base address
Base[LA]
• The address of any element of LA can be
computed using the formula
LOC (LA[K])=Base[LA]+W(K-LB)
• Where W is the number of words per
memory cell of the LA.
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Computer Memory
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Representation of Arrays in Memory
• Consider the AUTO array example, which records
number of automobiles sold each years from 1932 to
1984.
• Suppose AUTO appears in the memory as pictured in
the Figure.
Base(AUTO)=200, and W=4 words per cell.
LOC(AUTO[1932])= 200, LOC(AUTO[1932])= 204, …
• The address of array element for the year 1965 is
LOC(AUTO[1965])= Base(AUTO)+W(1965-LB)
= 200+4(1965-1932)
= 200+4(33)
= 200+132
= 332
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Array Characteristics
• Homogeneity
o All elements in an array must have the same data
type
• Contiguous Memory
o Array elements (or their references) are stored in
contiguous/consecutive memory locations
• Direct access to an element
o Index reference is used to access it directly
• Static data structure
o An array cannot grow or shrink during program
execution…the size is fixed
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ARRAY
Operations
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Array Operations
• Traversing: Accessing and processing each
element of ana array exactly once
o Display all contents of an array
o Count the number of elements of an array with a
given property
/* (Traversing a linear array)
Here LA is a linear array with lower bound LB and
upper bound UB. This algorithm traverses LA by
applying an operation PROCESS to each element of LA
*/
for (K=LB;K<=UB; K++)
PROCESS (LA[K])
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Array Operations (Traversing)
• Consider the AUTO array example, which records
number of automobiles sold each years from 1932 to
1984. Find the number NUM of years during which
more than 300 cars were sold.
NUM=0
for (K=LB;K<=UB; K++)
if AUTO[K]> 300
NUM=NUM+1
• Print year and number of automobiles sold in that
year.
for (K=LB;K<=UB; K++)
cout<< K<“\t” AUTO[K]
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Array Operations
• Insertion: Add an element at a certain
index
oWhat if we want to add an element at the
beginning?
• This would be a very slow operation! Why?
o Because we would have to shift ALL other elements over
one position
• What if we add an element at the end?
o It would be FAST. Why? No need to shift.
Prof. Dr. Wajid Aziz
Array Operations (Traversing)
• Print year and number of automobiles sold
in that year.
for (K=LB;K<=UB; K++)
cout<< K<<“\t”<<AUTO[K]
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Array Operations: Insertion
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• Algorithm: The algorithm inserts a data elements ITEM into Kth
position in a linear array LA comprising of N elements.
• The first four steps create a space in LA by moving downward
one location each element from Kth position on.
INSERT (LA, N, K, ITEM)
1. Set J=N
2. Repeat step 3 and 4 while J>=K
3. Set LA[J+1]=LA[J]
4. Set J=J-1
5. LA[K]=item
6. Set N=N+1
7. Exit
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Array Operations: Insertion
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• C++ Code Segment to Insert an item in an array: The
code segment inserts a data elements ITEM into Kth
position in a linear array LA comprising of N elements.
VOID INSERT (int LA[], int N,int K, int ITEM)
{
for (J =N ; J >= K; J--)
LA[J] = LA[J-1];
LA[K] = ITEM;
N=N+1
}
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• Algorithm: The algorithm deletes Kth element
from a linear array LA and assigns it to a variable
ITEM.
DELETE (LA, N, K, ITEM)
1. Set ITEM=LA[K]
2. Repeat for J=K to N-1
Set LA[J]=LA[J+1]
3. N=N-1
4. Exit
Prof. Dr. Wajid Aziz
• Deletion: Remove an element at a certain
index
oRemove an element at the beginning of the
array
• Performance is again very slow.
o Because ALL elements need to shift one position
backwards
oRemove an element at the end of an array
• Very fast because of no shifting needed
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Array Operations: Deletion
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Array Operations
Array Operations: Deletion
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• C++ Code Segment: The code deletes Kth element
from a linear array LA and assigns it to a variable
ITEM.
void DELETE (int LA[], int N, int K, int ITEM)
{
ITEM=LA[K]
for (J=K;J<N; J--);
LA[J]=LA[J+1];
N=N-1
)
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Array Operations
• Searching refers to the operation of finding
LOC of ITEM in an array.
• Searching through the array:
• Depends on the algorithm
• Some algorithms are faster than others
o More detail coming soon!
oLinear Search
oBinary Search
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Array Operations
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• Linear Search: It is a sequential searching algorithm where
o Start from the leftmost element of the array and one by one
compare ITEM with each element of array
o If ITEM matches with an element, return the index.
o If ITEM doesn’t match with any of elements, return -1..
• It is the simplest searching algorithm.
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Linear Search
Array Operations
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• Linear Search Code Segment: The code returns location of the
values
o If found it will return Index of the ITEM
o Otherwise, it will return -1
int search(int LA[], int N, int ITEM)
{
int J;
for (J = 0; J < N; J++)
if (LA[J] == ITEM)
return J;
return -1;
}
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Analysis of Linear Search
• Applicable for all data i.e. Sorted or
Unsorted
oBasic operation is “comparison”
oThey ONLY way to be sure that a value isn’t in
the array is to look at every single spot of the
array
oIf we have 100 elements then we have to make
100 comparisons to be sure about the value
oTherefore for “n” elements, the number of
comparisons will be “n” i.e. O(n)
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Binary Search
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Array Operations
• Binary Search: It is a searching algorithm for finding an
element's position in a sorted array.
• In this approach, the element is always searched in the
middle of a portion of an array.
• Example: Number Guessing Game from childhood
o I have a secret number between 1 and 100.
o Make a guess and I’ll tell you whether your guess is too high
or too low.
o Then you guess again and the process continues until you
guess the correct number.
o Your job is to MINIMIZE the number of guesses you make.
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Binary Search (Introduction)
• Number Guessing Game from childhood
o What is the first guess of most people?
• 50.
o Why?
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• No matter the response (too high or too low), the most
number of possible values for your remaining search is
50 (either from 1-49 or 51-100)
• Any other first guess results in the risk that the possible
remaining values is greater than 50.
o Example: you guess 75
o I respond: too high
o So now you have to guess between 1 and 74
» 74 values to guess from instead of 50
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Binary Search (Introduction)
• Applicable only on Sorted array
index
value
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• We are searching for the value, 19
• So where is halfway between?
o One guess would be to look at 2 and 118 and take
their average (60).
o But 60 isn’t even in the list
o And if we look at the number closest to 60
• It is almost at the end of the array
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Binary Search (Introduction)
• We quickly realize that if we want to adapt the
number guessing game strategy to searching
an array, we MUST search in the middle INDEX
of the array.
index
value
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o Index 4 stores 33
• The answer would be “less than”
• So we would modify our search range to in between index 0
and index 3
o Note that index 4 is no longer in the search space
o The second index we’d look at is index 1, since (0+3)/2=1
o Then we’d finally get to index 2, since (2+3)/2 = 2
o And at index 2, we would find the value, 19, in the array
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Array Operation
• We would ask, “is the number I am searching for, 19, greater or
less than the number stored in index 4?
• We then continue this process
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Binary Search (Introduction)
index
value
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19
o The lowest index is 0
o The highest index is 8
o So the middle index is 4
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• In this case:
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• Binary Search Code Segment: The code returns location of
the values
o If found it will return Index of the ITEM
o Otherwise, it will return -1
int search(int LA[], int Low, int High int ITEM)
{
int Mid;
Mid = (Low + High)/2
if (ITEM == LA[Mid])
return Mid
else if (ITEM > LA[Mid])
Low = Mid + 1
else
High = Mid - 1
}
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Array Operation
Analysis of Binary Search
• Let’s analyze how many comparisons (guesses) are
necessary when running this algorithm on an array
of n items
First, let’s try n = 128
o After 1 guess, we have 64 items left,
o After 2 guesses, we have 32 items left,
o After 3 guesses, we have 16 items left,
o After 4 guesses, we have 8 items left,
o After 5 guesses, we have 4 items left,
o After 6 guesses, we have 2 item left
o After 7 guesses, we have 1 items left.
o After 8 guesses, we have 0 items left.
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Analysis of Binary Search
• General case for n items
oAfter 1 guesses, we have n/2 items left,
oAfter 2 guesses, we have n/4 items left,
oAfter 3 guesses, we have n/8items left,
oAfter 4 guesses, we have n/16 items left,
o…………………
o…………………
o…………………
oSo on until we have 1 item left
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Analysis of Binary Search
• General case for n items
oAfter 1 guesses, we have n/2 items left,
oAfter 2 guesses, we have n/4 items left,
oAfter 3 guesses, we have n/8items left,
oAfter 4 guesses, we have n/16 items left,
o…………………
oAfter 10 guesses, we have
oAfter k guesses, we have
oWe will stop when we left with 1 item
Prof. Dr. Wajid Aziz
n/21
n/22
n/23
n/24
n/210
n/2k
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Analysis of Binary Search
• So we will stop once
n
1
2k
n  2k
k  log2 n
• This means that a binary search
roughly takes log2n comparisons when
searching in a sorted array of n items
• Efficiency of Binary Search is O(log2n)
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Linear Search vs Binary Search
• Linear search O(n)
• Binary Search O(log2n)
• Binary Search is more efficient
n
log n
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1024
65536
1048576
33554432
1073741824
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Multi-Dimensional Array
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• Most of the programming languages allow
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• A two-dimensional m×n array A in memory is a
collection of m.n data elements such that each
element is specified by a pair of integers (J, K)
called subscripts, with the property
1≤J≤m
and
1≤K≤n
• The elements of A with first subscript J and
second subscript K are denoted by
A[J, K] or
Prof. Dr. Wajid Aziz
o Two-dimensional arrays - having 2 subscripts
o Three-dimensional arrays-having 3 subscripts
• Some of the programming languages allow
number of dimensions for an array as high as
7.
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Two-Dimensional Array
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• A multi-dimensional array can be termed as an array
of arrays that stores homogeneous data in tabular
form.
A[J][K]
Two-Dimensional Array
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• A two-dimensional m×n
array
A
will
be
represented in memory
by a block of m.n
sequential
memory
locations .
• Specially programming
language will store the
array in
o Column-major
order
(column by column)
o Row-major order (row by
row)
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Two-Dimensional Array
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• Like one-dimensional array, the computer keeps the
track of Base(A)— the address of first element
A[1][1].
• To compute address LOC(A[J][K]), use the formula for
column-major order
o LOC(A[J][K])=Base(A) + w(M(K-1)+(J-1))
• To compute address LOC(A[J][K]), use the formula for
row-major order
o LOC(A[J][K])=Base(A) + w(N(J-1)+(K-1))
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Two-Dimensional Array
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• Suppose Base(score)=100 and w=4 words per
memory cell.
• The LOC(score[10][ 3]) using row-major order is
LOC(Score[J][K])=Base(A) + w(N(J-1)+(K-1))
LOC(Score[10][3])=100 + 4(4(10-1)+(3-1))=100+4(36+2)=252
• The LOC(score[10][ 3]) using colum-major order is
LOC(Score[J][K])=Base(A) + w(M(K-1)+(J-1))
LOC(Score[10][3])=100 + 4(25(3-1)+(10-1))=100+4(50+9)=336
Prof. Dr. Wajid Aziz
Students
Score1
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Score3
Score4
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Two-Dimensional Array
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• Suppose 25 students in a class are giving 4 tests as
shown in following Table.
• Assume students are numbered from 1 to 25.
Two-Dimensional Array
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• Two dimensional arrays are called matrices in
mathematics and tables in business applications.
• A two-dimensional 3×3 array A in C++ is shown
below:
𝑨 𝟎 𝟎 𝑨 𝟎 𝟏 𝑨[𝟎][𝟐]
𝑨 𝟏 𝟎 𝑨 𝟏 𝟏 𝑨[𝟐][𝟐]
𝑨 𝟐 𝟎 𝑨 𝟐 𝟏 𝑨[𝟐][𝟐]
• The number of elements in 3×3 array A is 9.
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General Multi-Dimensional Arrays
General Multi-Dimensional Arrays
• The n-dimensional m1×m2×m3×…×mn array B is a
collection of m1.m2.m3…mn data elements in which
each element is specified by a list of integers— such
as K1,K2,K3,…,Kn called subscripts, with property that
• Suppose B is a three-dimensional 2×4×3 array.
• The data elements of array B are 2.4.3=24.
• These 24 elements of B appear in three layers called page.
o 1 ≤ K1 ≤ m1’
1 ≤ K2 ≤ m2 ’ ….’ 1 ≤ Kn ≤ mn
• The elements of B with K1,K2,…,Kn will be denoted by
o B[K1][K2]…[Kn]
• Specifically, programming language will store the
array B in
o Column-major order
o Row-major order
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o Each page consist of 2×4 rectangular array of elements with same third
subscript.
o Three subscripts of an elements of 3-dimensional array are called row,
column and page.
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General Multi-Dimensional Arrays
General Multi-Dimensional Arrays
• The two ways of storing the 3-dimensional arrays
are shown in the following figure..
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• For a given subscript Ki, the effective index Ei of
length Li is the number of indices preceding Ki in
the index set and is calculated as
Ei=Ki-LB
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• The address LOC(C[K1][K2], …, [KN]) of an arbitrary
element of array C can be obtained using column
major order as
Base(C)+w((((…(ENLN-1+EN-1)LN-2+…+E3)L2+E2)L1+E1)
• By row-major order
Base(C)+w((((…(E1L2+E2)L3+…+E3)L4+…+EN-1)LN+EN)
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General Multi-Dimensional Arrays
General Multi-Dimensional Arrays
• Suppose a three-dimensional array M is declared
using
M(2:7, -4:1, 5:9)
• The lengths of three dimensions of M are
o L1=7-2+1=6
o L2=1-(-4)+1=6
o L3=9-5+1=5
• The number of elements M are
o Elements=L1.L2.L3=6.6.5=180
• Effective indices of M[5][-1][7]
o E1=5-2=3
o E2=-1-(-4)=3
o E3=7-5=2
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• The location of M[5][-1][7] using row-major order
is
LOC(M[5][-1][7])=Base(M)+w((E1L2+E2)L3+E3)
E1L2=3.6=18
E1L2+E2=18+3=21
(E1L2+E2)L3=21.5=105
(E1L2+E2)L3+E3 =105+2=107
Let Base(M)=200 and w=4
LOC(M[5][-1][7])=200+4(107)=200+428=628
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General Multi-Dimensional Arrays
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• The location of M[5][-1][7] using column-major
order is
LOC(M[5][-1][7])=Base(M)+w((E3L2+E2)L1+E1)
E3L2=2.6=12
E3L2+E2=12+3=15
(E3L2+E2)L1=15.6=90
(E3L2+E2)L1+E1=90+3=93
Let Base(M)=200 and w=4
LOC(M[5][-1][7])=200+4(93)=200+372=572
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