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

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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

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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) A

, A

,….., A

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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:

A:ARRAY [1 ………10] of real;

• C/C++ Language:

Syntax:

Data type array name [size];

Example: int a[10];

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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.

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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)

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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

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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

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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;

}

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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

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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

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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.

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• 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

8 9

15 7 8

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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

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Procedure

delete (A, pos, Ub, item) {

item = A[pos];

for i = pos to Ub-1 do

A[i] = A[i+1];

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.

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 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

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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

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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;

} }

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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.

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

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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’’;

}

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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

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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) }

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