String Basics and Memory Representation in DSA Typescript - Build from Scratch

String Basics and Memory Representation

📖 Scenario: Imagine you are working with a simple text editor that stores and manipulates words. Understanding how strings are stored and accessed in memory helps you manage text efficiently.

🎯 Goal: You will create a string variable, check its length, access characters by their position, and print the results to understand how strings work in memory.

📋 What You'll Learn

Create a string variable with a specific word

Create a variable to store the length of the string

Access a character at a specific position in the string

Print the string, its length, and the accessed character

💡 Why This Matters

🌍 Real World

Understanding strings and how they are stored helps in text processing, searching, and editing tasks in software like text editors, messaging apps, and data parsers.

💼 Career

String manipulation is a fundamental skill for software developers, data analysts, and anyone working with text data or user input.

1

Create a string variable

Create a string variable called word and set it to the exact value "hello".

DSA Typescript

// Your code here

Hint

Use const to create a variable and assign the string "hello".

2

Find the length of the string

Create a variable called length and set it to the length of the string word using word.length.

DSA Typescript

const word: string = "hello";
// Your code here

Hint

Use word.length to get the number of characters in the string.

3

Access a character by position

Create a variable called firstChar and set it to the first character of word using word[0].

DSA Typescript

const word: string = "hello";
const length: number = word.length;
// Your code here

Hint

Use word[0] to get the first character of the string.

4

Print the string, its length, and first character

Print the variables word, length, and firstChar each on a new line using console.log.

DSA Typescript

const word: string = "hello";
const length: number = word.length;
const firstChar: string = word[0];
// Your code here

Hint

Use three console.log statements to print each variable on its own line.

Practice

1. What does the term immutable mean when talking about strings in TypeScript?

easy

A. Once created, the string cannot be changed.

B. The string can be changed anytime.

C. The string is stored in a mutable array.

D. The string is stored as a number.

Solution

1. Step 1: Understand string immutability

   Strings in TypeScript cannot be changed after creation. Any modification creates a new string.

2. Step 2: Compare options with definition

   Only Once created, the string cannot be changed. correctly states that strings cannot be changed once created.

3. Final Answer:

   Once created, the string cannot be changed. -> Option A

4. Quick Check:

   Immutable means unchangeable = A [OK]

Hint: Remember: strings can't be changed after creation [OK]

Common Mistakes:

• Thinking strings can be changed like arrays
• Confusing mutable and immutable
• Assuming strings are stored as numbers

2. Which of the following is the correct way to declare a string variable in TypeScript?

easy

A. let name: number = 'Alice';

B. let name = 123;

C. let name: string = 'Alice';

D. let name: string = 123;

Solution

1. Step 1: Check variable declaration syntax

   TypeScript uses let variableName: type = value; syntax.

2. Step 2: Match type and value

   let name: string = 'Alice'; declares a string variable with a string value correctly.

3. Final Answer:

   let name: string = 'Alice'; -> Option C

4. Quick Check:

   Correct string declaration = A [OK]

Hint: Match variable type with value type exactly [OK]

Common Mistakes:

• Assigning number to string type
• Missing type annotation
• Using wrong syntax for declaration

3. What will be the output of the following TypeScript code?

let greeting = 'Hello';
greeting[0] = 'Y';
console.log(greeting);

medium

A. Yello

B. undefined

C. Error: Cannot assign to string index

D. Hello

Solution

1. Step 1: Understand string immutability in TypeScript

   Strings cannot be changed by index assignment; such operations do not modify the string.

2. Step 2: Analyze the code behavior

   Assigning greeting[0] = 'Y' does nothing; greeting remains 'Hello'.

3. Final Answer:

   Hello -> Option D

4. Quick Check:

   Strings immutable, no change = Hello [OK]

Hint: Strings can't be changed by index assignment [OK]

Common Mistakes:

• Expecting first letter to change
• Thinking string is mutable like array
• Assuming error will be thrown

4. What best describes the result of the following TypeScript code snippet:

let str: string = 'Test';
str[1] = 'a';
console.log(str);

medium

A. No error, output is 'Test'

B. TypeError: Cannot assign to string index

C. SyntaxError: Invalid assignment

D. ReferenceError: str is not defined

Solution

1. Step 1: Check string mutability rules

   Strings are immutable; assigning to an index does nothing.

2. Step 2: Analyze the code behavior

   str[1] = 'a' does nothing; str remains 'Test', console.log outputs 'Test'. No error.

3. Final Answer:

   No error, output is 'Test' -> Option A

4. Quick Check:

   Strings immutable, index assignment ignored = 'Test' [OK]

Hint: Strings cannot be changed by index assignment [OK]

Common Mistakes:

• Assuming no error and string changes
• Confusing syntax error with runtime error
• Expecting ReferenceError

5. Given the following code, what will be the output?

let s1 = 'abc';
let s2 = s1 + 'd';
console.log(s1);
console.log(s2);

Choose the correct output considering string immutability and memory representation.

hard

A. abcabc

B. abcabcd

C. abcdabcd

D. Error at runtime

Solution

1. Step 1: Understand string concatenation and immutability

   Concatenating strings creates a new string; original strings remain unchanged.

2. Step 2: Trace variable values after operations

   s1 remains 'abc'; s2 becomes 'abcd'.

3. Final Answer:

   abc
   abcd -> Option B

4. Quick Check:

   Original unchanged, new string created = abc and abcd [OK]

Hint: Concatenation creates new string; originals stay same [OK]

Common Mistakes:

• Expecting s1 to change after concatenation
• Thinking concatenation modifies original string
• Assuming runtime error on concatenation