LECTURE 02. Data Structures And Algorithms

Data Structures And Algorithms

LECTURE 02

Course Outline:

 Arrays and Records:

- Arrays

- Traversing a Linear Array - Inserting into Linear Array - Deleting from Linear Array

 Application of Linear Array:

- Bubble Sort - Linear Search - Binary Search

- Two Dimensional Arrays - String Handling Methods

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Fundamentals of Data Structures & Algorithms

ARRAYS

 Storage of data into memory can be done linearly or non-linearly.

 Linear – data will be organized in a sequential or linear order.

 Non-linear – data is organized in a non-linear order.

 The categorization of linear methods are:

- Arrays

- Linked list



In arrays the data will be stored in continuous memory location whereas,

- in linked list the data will be stored in different address (memory) location.



The categorization of non-linear methods are:

- Graphs

- Trees

Linear Array

 It is an array.

 Linear array can be defined using a single letter or many letter shown below:

Example: A or AB

Index represents: 1,……, n

Operations on an Array

 Retrieval of an element: Given an array index, retrieve the corresponding value.

- If the array is sorted, this enables one to compute the minimum, maximum, median (or in general, the ith smallest element)

essentially for free in O(1) time.

 Search: Given an element value, determine whether it is present in the array.

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Length or Size of Array

 The length or the size ( N) of any array can be found using:

N = Ub – Lb + 1 where Lb – Lower index Ub – Higher index

Example 1:

a[10]

n = 10 – 1 + 1 = 9 + 1

= 10

 Example 2:

If LB = 55, UB = 97, then n = (97-55)+1=43

 Array can be represented by:

i) A(1), A(2), ….. , A(n) or ii) A[1], A[2], ……, A[n] or iii) A₁ , A₂ ,….., A_n

Declaration of Linear Array in various Languages

 The following are the various declarations of linear array in four languages (Pascal, Fortran and C/C++):

 FORTRAN Syntax:

Data type array name (size) Example: REAL A(10)

 PASCAL Syntax:

Array Name: ARRAY [start index….end index] of data type;

Example:

 C/C++ Language:

Syntax:

Data type array name [size];

Example: int a[10];

Representation of Linear Array in Memory

 Memory consists of collection of cells.

 Each cell is represented address location.

In an array a value is stored at ‘α’ this is called base address which is the ‘’address of lower bound’’. The next value would be stored at α+1, α+2……… etc.

0 1 2 3 4 . . N Addresses

storage cell

 Using the base address it is easy to find out value of the address holding a data item.

Loc (A[λ]) = Base address (A) + w * (λ - Lb)

where

 A = Array name

 A[λ] = λth element of Array A

 Base Address (A) = address of the first element of the array,

 w = number of words in a memory cell

 Lb = lower bound (start index)

Calculation of the address in an interval address

space in Memory

Example 1:

Base Address (A) = 1000

w = one (or word) = 1; LB = 1

Then,

Loc ((A[20]) = base address (A)+(1*(20-1)) = 1000 + (1*(20-1))

= 1019

Example 2:

int A[10] = {5, 6, 12, 4, 15, 45, 87, 1, 9, 13}

Loc((A[5]) = 100 + 4 * (5 – 0) = 120

Traversing a Linear Array



The method of visiting each element in an array at exactly once is called Traversing.



Traversing is used to perform one of the following operations:

- To read an element

- To print an element

- To process an element

 The following steps are used to traverse a linear array.

Steps:

1. Set index λ to Lb (lower bound) 2. Visit the value of λ

3. Increment the value of λ then compare this with Ub (upper bound)

4. If λ < Ub or λ = Ub then go to step 2.

5. Exit operation

Procedure

Traverse (A, Lb, Ub) {

for λ = Lb to Ub visit (A[λ]);

}

Example:

Write a procedure to find sum of n-numbers using array.

Sum (A, Lb, Ub) {

sum = 0;

For λ = Lb to Ub do sum = sum + A[λ];

Print ‘’sum=‘’, sum;

}

Insertion in Linear Array

 Example:

Insert the new element 17 to the next position of A[0]

Before Insertion After Insertion

15 7

8 9

18

15 17 7 8 9 18 150

151 152 153 154

A[0]

A[1]

A[2]

A[3]

A[4]

150 151 152 153 154 155

A[4]

A[3]

A[2]

A[1]

A[0]

A[5]

LB = 0; UB = 4

LB = 0; UB = 5

Figure: Insertion Operation

Insertion in Linear Array Algorithm

 Insertion of a new element into a linear array can be done at the end or at the middle of the array (or any place within the linear array).

 The following are the steps to be traversed during insertion operation.

Steps to insert a new element at ‘k’ in an array ‘A’

(i) Compare ‘’k’’ with Ub

where k = the place at which element is going to be inserted Ub = upper bound

(ii) If k = Ub+1, insert the new element at A[k], i.e. A[k] = new element

(iii) If k <= Ub then

a) shift A[Ub], A[Ub-1],…., A[k] to A[Ub+1], A[Ub]…… A[k+1]

b) Insert the new element at A[k]; A[k] = new element

(iv) Increment the values of Ub and n.

i.e. Ub = Ub+ 1 and n = n+1.



Procedure

Insert (A, Lb, Ub, pos, item) //A-array Name {

for j = Ub to pos do A[j+1] = A[j];

A[pos]= item; //new item inserted

Ub = Ub+1; //upper bound modified }

Deletion from Linear Array

Example: Delete element 9 next of position A[2] from the linear array.

Before Deletion After Deletion

15 7 8 9

15 7 8 A[0]

A[1]

A[2]

A[3]

A[0]

A[1]

A[2]

150 151 152 153

150 151 152

LB = 0; UB = 3

LB = 0; UB = 2

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Deletion from Linear Array

 This method is used to delete an element from linear array.

 The following steps are traversed to delete an item at the kth place.

 Steps to delete an element from kth place on an array A are:

(i) Copy the element A[k] into a variable called item.

item = A[k]

(ii) Shift the values A[k+1], A[k+2], … , A[Ub] to A[k], A[k+1], … , A[Ub-1]

(iii) Decrement the values of Ub and n.

Ub = Ub -1 and n = n -1

Deletion from Linear Array

Procedure

delete (A, pos, Ub, item) {

item = A[pos]; //delete item

for i = pos to Ub-1 do

A[i] = A[i+1]; //shift the data

Ub = Ub-1;

}

Sorting – Bubble Sort Algorithm

 Consider an array A, which has some elements. To display the elements in ascending order scan the array ‘A’ for n-1 times.

 Scan-1

i) Compare the first element with the second element of the array. If the first element is smaller than second element then no need of interchange process. Otherwise swap the elements.

ii) Compare the second element with the third one. If second element is smaller than the third element then need not interchange. Otherwise interchange the elements.

iii) Do this process until the companion of (n-1)th element with nth element.

iv) At the end of the first scan, the biggest element will be moved to the nth position of an array A.

 Scan-2

i) Compare the first element with second element. If it is smaller need not swap it.

Otherwise swap the elements.

ii) Then compare the second element with third element. If small, need not interchange. Otherwise elements have to be swapped.

iii) Do this process until (n-2)th element is compared with (n-1)th element.

iv) At the end of second scan, the second biggest element will be moved to (n-1)th place.

 Scan (n-1)

Repeat the above steps until (n-1)th scan

Example:



Consider the following array:

a[1]=43, a[2]=72, a[3]=10, a[4]=23, a[5]=1



Sort the values in ascending order.

Refer to the next slide for the solution.

Bubble Sort

Let i – is a pointer which counts the number of scan.

j – is a pointer which is used to compare the first element and second element.

Refer to ‘’previous slide’’ for the solution of the example.

Procedure (a, n)

for i = 1 to n-1 do

for j = 1 to n-i do //number of scan if (a[j] > a[j+1]) then

{

temp = a[j];

a[j] = a[j+1];

a[j+1] = temp;

} }

Appah Bremang 29

Linear Search

 Consider an array ‘A’ with ‘n’ elements. An element X is to be search from the array A. If the element is found then the search succeeds else the search fails.

 The following are the steps to be tracked.



Steps involved in the Algorithm:

i) Compare X with A[1]. If X = A[1] then print as ‘’Search Success’’.

Otherwise compare X with A[2]. If X = A[2] print this as ‘’Search Success’’. Do this process until X is found in A[n].

ii) If no element is equal to X, then print this as ‘’Search failed’’

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 Example: Consider an array with the following elements (10).

The query is to search the element 25 from the given array.

10 7 13

9 25 17 15 6 19

5 A[1]

A[2]

A[3]

A[4]

A[5]

A[6]

A[7]

A[8]

A[9]

A[10]

Pass 1: 10 7 13 9 25 17 15 6 19 5 25 not matched

Pass 2: 10 7 13 9 25 17 15 6 19 5 25 not matched

Pass 3: 10 7 13 9 25 17 15 6 19 5 25 not matched

Pass 4: 10 7 13 9 25 17 15 6 19 5 25 not matched Pass 5: 10 7 13 9 25 17 15 6 19 5 25 matched

Procedure

Search (A, n, X) //X is the search data {

for i = 1 to n do if (X = A[i]) then {

Print ‘’Search Success’’;

Return;

}

Print ‘’Search Failed’’;

}

Binary Search



Consider a sorted array ‘A’ with ‘n’ elements. Assign the values lower bound as Lb and upper bound as Ub.



Example:

Query to search the element 60 from the following array:

A: [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

Stored Data 5 13 18 25 33 48 54 60 80 85

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 The query is to search the element 60 from the given array

A: [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

Stored data: 5 13 18 25 33 48 54 60 80 85 Search item: 60

Here Lb = 7 and Ub = 16

Pass 1

Compute mid = [(Lb + Ub)/2] = [(7 + 16)/2] = 11

A: [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

Data: 5 13 18 25 33 48 54 60 80 85

60 – search item is greater than the mid value Action – change the lower bound = 12

Pass 2

Compute mid = [((12+16)/2)] = 14

A: [7] [8] [9] [10] [11] [12] [13] [14] [15] [16]

Data: 5 13 18 25 33 48 54 60 80 85

60 success Appah Bremang 35

Binary Search Algorithm

 Consider a sorted array ‘A’ with ‘n’ elements. Assign the values lower bound as Lb and upper bound as Ub.

 Using the following steps (or algorithm) it is possible to find out the element ‘X’ from A.



Steps (or algorithm):

i) Calculate mid value as mid = (Lb + Ub)/2.

ii) Compare X with A[mid]. If X = A[mid] then X is found.

iii) If X < A[mid] repeat step(i) and step(ii) from 1 to mid-1.

iv) If X > A[mid] repeat step(i) and step(ii) from mid+1 to n.

v) If no value is equal to X then return (0).

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

Procedure

Binary Search (a, n, X, Ub, Lb) {

while (Lb <= Ub) do {

mid = (Lb + Ub)/2 if (X = A[mid]) then return (mid);

else

if (X < A[mid]) then Ub = mid-1;

else

Lb = mid+1 }

Return (0)

Two-Dimensional Arrays

 The method of representing data by row-wise and column-wise is called 2-Dimensional array.

 Syntax:

data type name[row size][column size];

 Range:

- The size of the row is called a row range.

- The size of column is called a column range.

Multiplication of row range and column range is equal to the maximum elements of 2-D array.

The maximum elements of 2-D array = row range * column range.

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 Consider the following array: int a[3][2].

 This array can hold a maximum of 6 elements in the memory.

Each location can be accessed using indexing or subscription.

column1 column2 Row1 a[1][1] a[1][2]

Row2 a[2][1] a[2][2]

Row3 a[3][1] a[3][2]

Representation of 2-Dimensional Array

 An example of 2D array can be arr[2][3] containing 2 rows and 3 columns.

 The arr[0][1] is an element placed at 0^th row and 1^st column in the array.

 A 2-dimensional array can thus be represented as given below:

Fig: Representation of 2D Array in Memory

Memory Representation of 2-D Array

 Two methods are used to represent the 2-dimensional arrays into memory.

i)

Row major representation.

ii)

Column major representation.

 In the row major method, the first row of the array ‘a’ will be stored in the first location. The second row of the array ‘a’ will be stored in the second location. So elements will be stored in row-wise.

 In the case of column major method, the elements will be stored in column-wise. First column elements will be stored first and then the second column element.

a[1][1]

a[1][2]

a[2][1]

a[2][2]

a[3][1]

a[3][2]

a[1][1]

a[2][1]

a[3][1]

a[1][2]

a[2][2]

a[3][2]

Row Major Column Major

Position 1

Position 2

Position 3

Position 1

Position 2

Memory Representation of Two-Dimensional Array

Implementation of 2-Dimensional Array

 In ´´C´´ language, the 2-Dimensional array can be represented as:

int a[rs][cs]

where a – name of the array rs – Row size

cs – Column size

After the declaration of this statement the compiler will assign, rs*cs continuous memory location in the memory and will assign a name ´a´ for this array.

Then the data will be stored in these address locations by row major method.

 Base Address

The address of a[0][0] is called the base address. Base (a) = a[0][0]. The dsize is the size of data in the memory. To find out the address a[0][1] of an element the following formula is used:

a[0][1] = base(a) + 1*dsize.

To find out the address a[0][2] of an element:

a[0][2] = base(a) + 2*dsize.

To find out address of the element which is stored at a[i][j]

the following method is used:

a[i][j] = base(a) + (i*cs*j)*dsize

Appah Bremang 44

Calculation of the Address of an element in the 2-D Array:

 Example:

int A[3][4] = {10, 13, 24, 3, 41, 5, 6, 17, 8, 91, 10, 16}

Each element of A is 4bytes long

This 2-D array will be presented externally and in the memory of the computer using ROW- MAJOR-ORDER

 ROW-MAJOR-ORDER

 Store as ROW-MAJOR-ORDER

int A[3][4] = {10, 13, 24, 3, 41, 5, 6, 17, 8, 91, 10, 16}

Here row = 3 and column = 4 or M = 3 and N = 4

Stored as COLUMN-MAJOR-ORDER

 Similarly, for the Column-Major-Order representation, let us consider the matrix.

int A[3][4] = {10, 13, 24, 3, 41, 5, 6, 17, 8, 91, 10, 16}

Here row = 3 and column = 4 or M = 3 and N = 4

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Low Addresses High Addresses

SUMMARY OF ROW AND COLUMN

MAJOR

Representation of tables using Arrays:

 Consider the following table player list:

Hockey Basketball Football Leo

Lydia Paul

Stephen Richard David

Samuel Binny David Peter Helena Lawal

This list is stored in the memory using different methods

 Method 1, Using 2D Array ^Leo

Lydia Paul

Stephen Richard David

Samuel Binny David Peter Helena

Lawal Hockey

Basketball

Football

The player list is stored in an array of

M * N or N * M size.

Where m – represents group (item) names and n – represents maximum members of the team

Draw Back:

- Memory is wasted

- To store this player list in the memory, we need 18 memory locations.

- Out of the 18 memory locations only 12 locations are used by this array.

- The remaining 6- places unused.

52

Method 2 – Using 1-Dimensional Array

Leo Lydia

Paul Stephen Richard David Samuel

Binny David Peter Helena

Lawal Hockey

Basketball

Football

Advantage:

The memory is not wasted.

Draw Back:

- Retrieving a player from a particular group is not possible in this method, because there are no separation between two groups.

Here the elements of player list are stored in a continuous memory locations using 1 – Dimensional array.

First group of the play list is stored in the starting location of memory.

The second group name is stored in second location of the memory.

Third group name is stored in third location.

Method 3 – Using 1-Dimensional Array

Here a player list is stored in a continuous memory location, but each team is separated by a ($) dollar sign.

By separating each group by a dollar symbol it is very easy to read a player name from any group

Reading starts from the starting location of the array.

Leo Lydia

Paul

$ Stephen Richard David

$ Samuel

Binny David Peter Helena

Lawal

54 $

String Handling Methods

 Strings are second category of data.

 The char array starts with 0 initialization. An example declaration is shown below:

Example: Char str[15]

 The above declaration tells the compiler that an array str is initialized with fifteen spaces.

 If the string ‘’Hello , World!’’ is stored in the array the representation is as follows:

 Characters: H e l l o , w o r l d !

 Subscripts: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 H e l l o , w o r l d !

There are different variations of string functions.

They are:

 Strlen()

 Strcpy()

 Strncpy()

 Strcat()

 Strncat()

 Strcmp()

 Strncmp()

strlen()

 As the name indicates this function calculates the length of the string. The following gives the syntax declaration of

strlen ().

 Syntax: len = strlen(ptr):

where len is an integer and ptr is a pointer to char.

This function returns the length of the input string excluding the NULL.

Example:

int len;

char str[15];

strcpy(str, ‘’Hello, world’’);

len = strlen(str);

 Strcpy()

This is a library function which copies one string to another string

Syntax: strcpy(ptr1, ptr2); where ptr1 & ptr2 are pointers to char.

E.g. char S[25];

char D[25];

strcpy(S,’’This is string1.’’); //Putting text into a string.

strcpy(D,S); //copying a whole string from S to D strcpy(D,S[8]); //copying the tail end of string S to D

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 Strcat()

This library function concatenates or combine two strings.

The syntax is given below:

Syntax:

strcat(ptr1, ptr2); where ptr1 and ptr2 are pointers to char

Example:

char S[25] = ‘’world!’’;

char D[25]=‘’Hello,’’;

strcat(D, S); // concatenating the whole string S onto D. // gives Helloworld.

 Strncat()

syntax:

strncat(ptr1, ptr2, n); where ptr1 and ptr2 are pointers to char

and n is an integer.

char S[25] = ‘’world!’’;

char D[25] = ‘’Hello,’’;

concatenating five characters from the beginning of S onto the end of D and placing a null at the end.

 Strncat(D, S, 5);

 Strncat(D, S, strlen(S)-1);

Both would result in D containing ‘’Hello,world’’.

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 Strcmp()

This library function compares two strings for equality.

syntax:

diff = strcmp(ptr1,ptr2);

where diff is an integer and ptr1 and ptr2 are pointers to char e.g.

char s1[25]=‘’pat’’;

char s2[25]=‘’pet’’;

diff = strcmp(s1,s2); //diff will have a negative value if s1<s2

diff = strcmp(s2,s1); //diff will have a positive value if s1>s2

diff = strcmp(s1,s1); //diff will have a value of zero(0) if s1=s2

 Strncmp

syntax: diff = strncmp(ptr1, ptr2, n);

where diff and n are integers, and ptr1 and ptr2 are pointers to char.

Strncmp() is used to compare the first n characters of two strings. The strings are compared character by character starting at the characters pointed at by the two pointers.

if the first n strings are identical, the integer value zero (0) is returned.

Example ‘C’ programs for string handling methods

#include<stdio.h>

#include<string.h>

void main() {

char s[20] = ´´kwame´´;

char t[10] = ´´Ansah´´;

clrscr();

strcat (s, t);

printf(´´the new string is %s´´,s) getch()

}

Output: the new string is kwameAnsah

#include<stdio.h>

#include<string.h>

void main() {

char s[20] = ´´kwame´´;

char t[10] = ´´Ansah´´;

clrscr();

strncat (s, t,3);

printf(´´the new string is %s´´,s);

getch() }

Output: the new string is kwameAns

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main() {

char alpha[ ] = ´´kwame´´;

clrscr();

printf(´The string %s contains %d character´´,alpha, strlen(alpha));

getch();

}

Output: The string kwame contains 5 characters