Exercises to Learn Algorithm and Programming

Exercises to Learn Algorithm Programming| Pascal| Python| ICT O/L| ICT A/L

Problem 1
Write an algorithm to read two numbers and print their sum.

Algorithm (Pseudocode>

Start
Read A, B
SUM ← A + B
Print SUM
Stop

Pascal Program
program SumOfTwoNumbers;
var
  A, B, SUM: integer;
begin
  writeln('Enter value of A: ');
  readln(A);
  writeln('Enter value of B: ');
  readln(B);

  SUM := A + B;

  writeln('Sum = ', SUM);
end.

Python Program
# Read two numbers
A = int(input("Enter value of A: "))
B = int(input("Enter value of B: "))

# Calculate sum
SUM = A + B

# Print result
print("Sum =", SUM)

Problem 2
Write an algorithm to find the larger of two numbers.

Algorithm Pseudocode

Start
Read A, B
If A > B
	Print A
Else
	Print B
Stop


Pascal Program
program FindGreaterNumber;
var
  A, B : integer;
begin
  readln(A, B);

  if A > B then
    writeln(A)
  else
    writeln(B);
end.

Python Program
A = int(input())
B = int(input())

if A > B:
    print(A)
else:
    print(B)

Problem 3
Write an algorithm to check whether a number is even or odd

Algorithm Pseudocode

Start
Read N
If N mod 2 = 0
Print "Even"
Else
Print "Odd"
Stop


Pascal Program
program EvenOdd;
var
  N: integer;
begin
  writeln('Enter a number: ');
  readln(N);

  if N mod 2 = 0 then
    writeln('Even')
  else
    writeln('Odd');
end.

Python Program
N = int(input("Enter a number: "))

if N % 2 == 0:
    print("Even")
else:
    print("Odd")

Problem 4
Write an algorithm to calculate simple interest.
Formula:
SI = (P × R × T) / 100

Algorithm Pseudocode

Start
Read P, R, T
SI ← (P × R × T) / 100
Print SI
Stop


Pascal Program
program SimpleInterest;
var
  P, R, T, SI: real;
begin
  writeln('Enter Principal (P): ');
  readln(P);
  writeln('Enter Rate (R): ');
  readln(R);
  writeln('Enter Time (T): ');
  readln(T);

  SI := (P * R * T) / 100;

  writeln('Simple Interest = ', SI:8:2);
end.

Python Program
P = float(input("Enter Principal (P): "))
R = float(input("Enter Rate (R): "))
T = float(input("Enter Time (T): "))

SI = (P * R * T) / 100

print("Simple Interest =", SI)

Problem 5
Write an algorithm to reverse a number.

Algorithm Pseudocode

Start
Read N
REV ← 0
While N > 0
	DIGIT ← N mod 10
	REV ← REV × 10 + DIGIT
	N ← N / 10
EndWhile
Print REV
Stop


Pascal Program
program ReverseNumber;
var
  N, REV, DIGIT: integer;
begin
  writeln('Enter a number: ');
  readln(N);

  REV := 0;

  while N > 0 do
  begin
    DIGIT := N mod 10;
    REV := REV * 10 + DIGIT;
    N := N div 10;
  end;

  writeln('Reversed Number = ', REV);
end.

Python Program
N = int(input("Enter a number: "))
REV = 0

while N > 0:
    DIGIT = N % 10
    REV = REV * 10 + DIGIT
    N = N // 10

print("Reversed Number =", REV)

Problem 6
Write an algorithm to check whether a number is a palindrome.

Algorithm Pseudocode

Start
Read N
TEMP ← N
REV ← 0
While TEMP > 0
	DIGIT ← TEMP mod 10
	REV ← REV × 10 + DIGIT
	TEMP ← TEMP / 10
Endwhile
If REV = N
	Print "Palindrome"
Else
	Print "Not Palindrome"
Stop


Pascal Program
program PalindromeNumber;
var
  N, TEMP, REV, DIGIT: integer;
begin
  writeln('Enter a number: ');
  readln(N);

  TEMP := N;
  REV := 0;

  while TEMP > 0 do
  begin
    DIGIT := TEMP mod 10;
    REV := REV * 10 + DIGIT;
    TEMP := TEMP div 10;
  end;

  if REV = N then
    writeln('Palindrome')
  else
    writeln('Not Palindrome');
end.

Python Program
N = int(input("Enter a number: "))
TEMP = N
REV = 0

while TEMP > 0:
    DIGIT = TEMP % 10
    REV = REV * 10 + DIGIT
    TEMP = TEMP // 10

if REV == N:
    print("Palindrome")
else:
    print("Not Palindrome")

Problem 7
Write an algorithm to find the sum of digits of a number.

Algorithm Pseudocode

Start
Read N
SUM ← 0
While N > 0
	DIGIT ← N mod 10
	SUM ← SUM + DIGIT
	N ← N / 10
EndWhile
Print SUM
Stop


Pascal Program
program SumOfDigits;
var
  N, SUM, DIGIT: integer;
begin
  writeln('Enter a number: ');
  readln(N);

  SUM := 0;

  while N > 0 do
  begin
    DIGIT := N mod 10;
    SUM := SUM + DIGIT;
    N := N div 10;
  end;

  writeln('Sum of Digits = ', SUM);
end.

Python Program
N = int(input("Enter a number: "))
SUM = 0

while N > 0:
    DIGIT = N % 10
    SUM = SUM + DIGIT
    N = N // 10

print("Sum of Digits =", SUM)

Problem 8
Write an algorithm to count the number of digits in a number.

Algorithm Pseudocode

Start
Read N
COUNT ← 0
While N > 0
	COUNT ← COUNT + 1
	N ← N / 10
Endwhile
Print COUNT
Stop


Pascal Program
program CountDigits;
var
  N, COUNT: integer;
begin
  writeln('Enter a number: ');
  readln(N);

  COUNT := 0;

  while N > 0 do
  begin
    COUNT := COUNT + 1;
    N := N div 10;
  end;

  writeln('Number of digits = ', COUNT);
end.

Python Program
N = int(input("Enter a number: "))
COUNT = 0

while N > 0:
    COUNT = COUNT + 1
    N = N // 10

print("Number of digits =", COUNT)

Problem 9
Write an algorithm to check whether a year is a leap year.

Algorithm Pseudocode

Start
Read YEAR
If (YEAR mod 400 = 0)
Print "Leap Year"
Else if (YEAR mod 100 = 0)
Print "Not Leap Year"
Else if (YEAR mod 4 = 0)
Print "Leap Year"
Else
Print "Not Leap Year"
Stop


Pascal Program
program LeapYear;
var
  YEAR: integer;
begin
  writeln('Enter the year: ');
  readln(YEAR);

  if (YEAR mod 400 = 0) then
    writeln('Leap Year')
  else if (YEAR mod 100 = 0) then
    writeln('Not Leap Year')
  else if (YEAR mod 4 = 0) then
    writeln('Leap Year')
  else
    writeln('Not Leap Year');
end.

Python Program
YEAR = int(input("Enter the year: "))

if YEAR % 400 == 0:
    print("Leap Year")
elif YEAR % 100 == 0:
    print("Not Leap Year")
elif YEAR % 4 == 0:
    print("Leap Year")
else:
    print("Not Leap Year")

Problem 10
Write an algorithm to search an element in a list.

Algorithm Pseudocode

Start
Read N
Read array A
Read key
For i ← 1 to N
	If A[i] = key
		Print "Found"
		Stop
EndFor
Print "Not Found"
Stop


Pascal Program
program LinearSearch;
var
  A: array[1..100] of integer;
  N, i, key: integer;
  found: boolean;
begin
  writeln('Enter number of elements: ');
  readln(N);

  writeln('Enter array elements: ');
  for i := 1 to N do
    readln(A[i]);

  writeln('Enter key to search: ');
  readln(key);

  found := false;

  for i := 1 to N do
  begin
    if A[i] = key then
    begin
      writeln('Found');
      found := true;
      break;
    end;
  end;

  if not found then
    writeln('Not Found');
end.

Python Program
N = int(input("Enter number of elements: "))
A = []

print("Enter array elements:")
for i in range(N):
    A.append(int(input()))

key = int(input("Enter key to search: "))

found = False

for i in range(N):
    if A[i] == key:
        print("Found")
        found = True
        break

if not found:
    print("Not Found")

Programming Exercises for ICT A/L & ICT O/L

Function accept number return in words| Python| ICT A/L

Python function which accept number as input and return it in words

Exercises for Python| User defined function in Python| Function accept  number return in words| Exercises for Python List, String, functions and operators

ICT A/L

#function defined to accept two digit number as parameter and print the number in words

def digit1n2rword(num):
    ones = ["Zero", "One", "Two", "Three", "Four",
            "Five", "Six", "Seven", "Eight", "Nine"]

    teens = ["Ten", "Eleven", "Twelve", "Thirteen", "Fourteen",
             "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"]

    tens = ["", "", "Twenty", "Thirty", "Forty",
             "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"]

    if num < 10:
        return ones[num]
    elif num < 20:
        return teens[num - 10]
    else:
        return tens[num // 10] + (" " + ones[num % 10] if num % 10 != 0 else "")

#function defined to accept three digit number as parameter and print the number in words using the digit1n2rword

def digit3word(num):
    if num < 100:
        return digit1n2rword(num)

    hundreds = num // 100
    remainder = num % 100

    if remainder == 0:
        return digit1n2rword(hundreds) + " Hundred"
    else:
        return digit1n2rword(hundreds) + " Hundred " + digit1n2rword(remainder)

#function defined to accept any digit number as parameter and print the number in words using the function digit3word

def anyDigitWord(num):
    if num == 0:
        print("Zero")
        return

    place_values = ["", "Thousand", "Million", "Billion", "Trillion"]
    words = []

    i = 0
    while num > 0:
        part = num % 1000
        if part != 0:
            text = digit3word(part)
            if place_values[i]:
                text += " " + place_values[i]
            words.insert(0, text)
        num //= 1000
        i += 1

    print(" ".join(words))

Main Program which call the anyDigitWord function with given numbers

anyDigitWord(7)
anyDigitWord(19)
anyDigitWord(56)
anyDigitWord(105)
anyDigitWord(250)
anyDigitWord(123456786)

Output

Seven
Nineteen
Fifty Six
One Hundred Five
Two Hundred Fifty
One Hundred Twenty Three Million Four Hundred Fifty Six Thousand Seven Hundred Eighty Six

ICT AL Past Paper ListPhrases

Question 1

Match each of the given phrases relating to commerce with the most suitable item from the list below:

List = {advertising as a revenue model, credit-cards, Government e-Tendering services, Government to Citizen (G2C) service, group purchasing, harmful explosives, online market, payment gateway, perishable goods, social commerce, subscription as a revenue model, traditional marketplace}

Phrase	List Item
A place where buyers and sellers interact physically for exchange goods and services for a place.
These are usually prohibited to be sold or purchased through e-commerce systems
Users pay a regular fee to have full access to a website of a business
A subset of e-commerce that involves using social media to assist in the online buying and selling of products and services
Facilitates a payment transaction by the transfer of information between the e-commerce application and the back-end financial services providers through secure means
The renewal of vehicle revenue license using the Online Vehicle Revenue License Service offered by the relevant government office.

Question 2

In an emergency health problem where people have to stay at home for long period, the shops within the area remain closed for regular business activities. Under such circumstances the shops within a village or nearby town can help their community by practicing their business through e-commerce.

Consider the above scenario, fill the blanks in the following statements with suitable phrases from the given list.

List ={ advertising banners, an online marketplace, a shopping cart, a web product catalogue, cash-on-delivery, credit-cards, discount pricing, group purchasing, payment gateway, click and brick, pure clicks, subscription as a revenue model}

Phrase	List Item
In this emergency situation, shops follow the …………………….business model.
Shops must use ……………………… to allow customers to purchase more than one type of product in a single transactions.
The e-commerce site for cash shop can implement ………………….to display their products to the customers.
For business owners who cannot use payment gateway through online fund receipts and for the customers who do not have any online mode of payment can still be supported through ……………………..
------------------is one of the best ways to reduce the overhead costs of delivery within a local area such as a lane, street or housing scheme
The local shop owners can establish ………………….to serve their community better by enabling access to each shop’s services through a common portal.

Question 3

Fill the blanks in the following statements with suitable words from the given list of words.

List ={ Encryption, Copyright, Phishing, Plagiarism, Software Piracy}

Phrase	List Item
…………….helps to ensure the confidentiality of our data and information.
………………..is the attempt to acquire sensitive information by pretending as a trustworthy entity in an electronic communication.
The illegal copying, distribution, or use of software is known as ……………….and …………………….helps us to protect our software from such illegal use.

Question 4

Match each of the given phrases relating to computer networks with the most suitable item from the list below.

List ={ADSL Connection, DSL Connection, FTP, HTTP, Internet Layer, Malware, Phising, TCP, Tranport Layer, UDP}

Phrase	List Item
A simple and query based communication model with a minimum use of protocol mechanisms applied in trasport layer.
A protocol for data communication in the World Wide Web
The layer that defines the addressing and routing structures used for the TCP/IP protocol suite in the TCP/IP model.
The process of attepting to get sensitive information from smeone by pretending as a trustworthy person.
The connection that allows the data transmission at much greater speed and capacity than the narrowband services.

---

Important Notes:

Pure Brick-and-Mortar:

Traditional businesses with physical stores only and no online presence. Example: A local grocery store with no website.

Brick-and-Click:

Businesses that have both physical stores and an online presence. Example: Walmart or Target – customers can shop in-store or online.

Pure Click:

Businesses that operate only online, with no physical storefronts. Example: Amazon or Netflix.

Click-and-Mortar (Same as Brick-and-Click):

This is another term for Brick-and-Click businesses combining offline and online operations.

Memory Management in Operating Systems: Paging, Swapping & Beyond

Virtual memory

Virtual memory is a memory management technique that gives the illusion of a large, continuous memory space to programs, even if the physical memory (RAM) is limited. It allows the operating system to use disk space (e.g. a swap file or page file) to extend RAM, enabling multitasking and isolation between processes.

Physical memory

Physical memory refers to the actual hardware (RAM) installed in a computer. It's where data and programs are stored temporarily while they are being used or executed.

Pages and Frames

In operating system memory management, pages and frames are fundamental concepts related to paging, a memory management scheme that eliminates the need for contiguous allocation of physical memory. Pages and Frames are different aspects of memory.

Comparison of Pages and Frames

Feature	Pages	Frames
Definition	Fixed-size blocks of a logical (virtual) address space	Fixed-size blocks of physical memory (RAM)
Belong to	Process address space	Physical memory
Size	Same as frame size (e.g., 4KB, 8KB) to ensure 1-to-1 mapping	Same as page size
Purpose	Divide the virtual memory of a process into manageable units	Hold the actual data from pages loaded into physical memory
Managed by	Memory management unit (MMU) and OS	Memory management unit (MMU) and OS
Mapping	Pages are mapped to frames using page tables	Frames store the actual data of the pages mapped to them
Location	Exist in virtual memory (on disk or RAM)	Reside in physical memory (RAM)
Swapping	Pages can be swapped in and out of disk (e.g., to/from swap space)	Frames are occupied by pages when they are loaded into RAM
Example	A process has 10 pages in its virtual address space	The OS has 1000 frames available in RAM, each of 4KB

Comparison of MMU and Paging Tables

Aspect	Memory Management Unit (MMU)	Paging Tables (Page Tables)
Definition	Hardware component that handles memory address translation and protection.	Data structure used by OS to store mappings of virtual addresses to physical addresses.
Role	Performs address translation (virtual → physical) at runtime using page tables.	Stores the actual mappings and metadata for each page.
Location	Part of the CPU (hardware).	Stored in memory (RAM), managed by the OS.
Function	Translates virtual addresses to physical addresses using page table entries.	Keeps track of virtual pages, frame numbers, valid/invalid bits, etc.
Speed	Very fast; works in nanoseconds.	Slower access; must be read by MMU during address translation (may be cached in TLB).
Dependency	Depends on page tables for translation.	Used by the MMU to perform translation.
Managed by	Hardware (CPU manufacturers).	Operating System (software).

Note

MMU = The brain that performs the translation.
Page Table = The dictionary that contains the translations.

Virtual memory = Simulated memory using disk + RAM.
Physical memory = Real, hardware-based RAM.

Swapping

• Swapping is a memory management technique where entire processes are temporarily moved between main memory (RAM) and secondary storage (disk).
• Free up RAM when it's full, by moving inactive processes to disk (called "swap space"), and bringing them back when needed.
• Whole process is swapped in or out.

Segmentation

• Segmentation divides a process's memory into logical segments, such as code, data, stack, etc.
• Allows logical organization, protection, and sharing of memory.
• Each segment is of variable size, and has its own base and limit.

Comparision of Swapping and Segmentation

Feature	Swapping	Segmentation
Definition	Moving entire processes between RAM and disk	Dividing a process into variable-sized segments
Purpose	Manage memory when RAM is full	Allow logical division of process memory
Unit of Transfer	Entire process	Individual segments (e.g., code, data, stack)
Memory Division	Not divided logically — whole process is moved	Divided based on logical units of a program
Size	Process size can vary Segment sizes	Vary based on program structure
Addressing	Uses single linear address space per process	Each segment has its own base and limit
Use Case	Used in memory overcommitment (swap in/out)	Used for easier protection and sharing of memory
Overhead	High (due to entire process movement)	Moderate (due to managing multiple segments)

Note
Swapping = Moving entire processes in/out of memory.
Segmentation = Dividing process memory into logical parts.

A Beginner's Guide to File System in an Operating Systems

File Systems

A file system is a method to  store, organise, retrieve and manage data on storage devices (Hard disks, SSD, USB) by Operating system. There are many different file systems available. Key functions of file systems are: structure data into files and folders, tracks where data is physically stored on the disks, manage permissions and user access to files, supports creating, reading, writing, deleting and modifying files. These functions can be varied  file system to file system. One files system may not support with other file system. Diffrent operating systems uses different file systems.

Some of the widely used file systems are given below

1. FAT (File Alloction Table)
• FAT is the file systems introduced with Microsoft Disk Operating System (MS OS).
• FAT uses a File Allocation Table (FAT) to keep track of files in the storage devices.
• FAT and the root directory reside at a fixed location of the volume so that the system's boot files can be correctly located.
• To protect a volume, two copies of the FAT are kept.
• There are many variants such as Variants: FAT12, FAT16, FAT32
• The advantages of FATare simple, widely compatible
• The dis advantages of FAT are Limited size (4GB for FAT 32), It is not very secure
2. exFAT (Extended File Allocation Table)
• exFAT is used at USB drives, SD cards (cross-platform)
• Advantages are Supports large files, compatible with Windows & macOS
• Disadvantages are No journaling, less robust for system drives
3. NTFS
• (New Technology File System) is a proprietary file system developed by Microsoft. This is improvement of FAT. Windows (default)
• This improvement includes The capability to recover from some disk-related errors automatically, which FAT cannot.
• Support with Unicode encoding system
• Improved support for larger hard disks.
• Better security as permissions, journaling and encryptions are used to restrict access to specific files to approved users.
• Disadvantages Limited support on non-windows systems
4. ext (Extended File System)
• (Extended File System): Commonly used in Linux distributions, with versions like ext4 offering journaling and improved performance.
• There are many variant available ext2, ext3, ext4 in Linux
• Advantages are ext4 supports journaling, large files, and volumes
• Disadvantages are Limited compatibility with windows/macOs
5. HFS+ and APFS:
• Used by macOS, with APFS designed for solid-state drives, providing features like snapshots and encryption.
• Older macOs used HFS+ file systems and APFS is used by modern macOS
• Advatages of APFS are better performance, encryption and snapshots
• Disadvantages not natively writable on non-Apple systems

Partitioning of a Disk

Partitioning is the process of dividing a physical storage device into separate independent sections.These independent section is called partitions. This partitioning is used for hard disk or SSD. Each partition can have its own file system, operating system, or usage purpose.

Why Partition a Disk

There are many advantages of patiting a disk. Some of them are given below

• Organize Data – Keep system files separate from personal data.
• Dual Boot – Install multiple operating systems (e.g., Windows + Linux).
• Improve Performance – Manage storage more efficiently.
• Backup & Recovery – Isolate system and recovery partitions.
• Security – Limit access between partitions.

Partitioning Schemes

• MBR (Master Boot Record): Older system, supports up to 2 TB and 4 primary partitions.
• GPT (GUID Partition Table): Modern standard, supports larger disks and up to 128 partitions. Required for UEFI systems.

Types of Partitions

There are several types of Partitions are available

• Primary Partition: Bootable partition that can host an OS (max 4 primary partitions per disk in MBR).
• Extended Partition: A special type that holds multiple logical partitions (used when you need more than 4 partitions in MBR).
• Logical Partition: A sub-part of an extended partition. Useful for organizing data.
• EFI/System Partition: Required for booting in UEFI systems (in GPT partitioning).

Disk Formatting

Disk formatting is the process of preparing a storage device (like a hard drive or SSD) to store data.

There are two main types of disk formatting is available. They are Low-Level Formatting and High-Level Formatting.

Low-level formatting

Low-level formatting is the process of physically preparing the disk by creating sectors and tracks on the surface of the disk. Low-level formatting is done at the factory for modern drives. It defines how data will be physically stored on the disk. It is Rarely done by users today. It Lays down the magnetic markers for sectors/tracks.

High-Level Formatting

High-level formatting is the process of setting up a file system on the disk. Different Operating system can setup different file systems example windows can setup NTFS or FAT32 same time Linux can setup ext4. Operating system can store and manage files after the High-Level Formatting is done. Performed by the user (e.g., using tools like Windows Disk Management or mkfs in Linux or format in DOS). Tasks are done by High-Level formatting: Initializes the file system. Creates boot sector, file allocation table, root directory, etc. High-Level Formatting Can be quick format which just sets up file system structures, or full format which scans for bad sectors too.

Ex : Using DOS command; if we use to full formating c: drive   format c:   quick format c: drive  format C: /q  formating c: drive with DOS intalled format c:/s

Feature	Low-Level Formatting	High-Level Formatting
Level	Physical	Logical (file system)
Frequeny	Rarely done by users	Commonly done
Performed By	Manufacturer (mostly)	User or OS
Purpose	Create sectors/tracks	Create file system structure

The Different between Word Processing and Desktop Publishing

The Different between Word Processing and Desktop Publishing (DTP)

Word processing is creating, editing, formatting and printing text based documents using computer software. This software is called Word processing software. Examples for some of the word processing software are Micro soft word, Open office writer, Apple Pages, Google docs and Word perfect etc.

In other hand Desk top publishing is used to create high quality, visually rich printed or digital documents using software. This software is called desk top publishing software. Example for some of the desk top publishing software is Adobe InDesign, QuarkXPress, Scribus (Open source) and PageMaker.

Description	Word Processing	Desk top publishing
Use	Text-heavy documents	For professional-looking, visually complex layouts
Software	Microsoft Word, Google Docs, Open Office Writer, Apple Pages	Adobe InDesign, QuarkXPress, Scribus (Open source), PageMaker
Best For	Letters, Reports, Resumes	Magazines, Brochures, Newsletters

Desk Top Publishing (DTP)

Disk Fragmentation and Defragmentation

Disk Fragmentation

• Fragmentation is the unintentional division of Disk into many small free areas that cannot be used effectively due to scattered storage of file fragments.
• It Occurs when files are stored in non-contiguous sectors on a disk. File operations are dynamic nature, where files grow and shrink, and blocks of data become available after deletions.
• Fragmented files require head to move more for read or write operations, It increases access times and reduceds performance, mainly in hard disks

Defragmentation

• Defragmentation is a process that locates and eliminates file fragments by rearranging them.
• The defragmentation tool analyzes the disk, identifies fragmented files, and moves them so that each file occupies a single contiguous block of space. It also consolidates free space to minimize fragmentation in the future.
• Defragment a disk periodically is advisable, if it is heavily used.
• Modern operating systems often automatically handle fragmentation using inbuilt tool.
  Example:
  Windows: The "Defragment and Optimize Drives"
  Linux: While Linux filesystems are generally more resistant to fragmentation, tools like e4defrag can be used for ext4 filesystems.
  
• Solid-state drives (SSDs) do not require defragmentation. Its architecture allows for equal access times regardless of data location. Defragmenting an Solid-state drives(SSD) can reduce its lifespan due to unnecessary write cycles.

Benefits of Defragmentation

• Improved Performance:
  By reducing the time it takes for the disk to access files, defragmentation can lead to faster boot times, quicker file access, and overall improved system responsiveness.


• Extended Drive Life:
  Although the effect is more pronounced on Hard Disks, better performance can lead to less wear and tear over time.

Limitations of Defragmentation

• Time-Consuming:
  Depending on the size of the disk and the level of fragmentation, defragmentation can take a significant amount of time. 


• Temporary Impact:
  After defragmentation, further file operations may lead to fragmentation again, necessitating ongoing maintenance.