String Basics and Memory Representation in DSA Typescript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to get the length of the string.

DSA Typescript

const str = "hello";
const length = str.[1];

Drag options to blanks, or click blank then click option'

Alength

Bsize

Ccount

Dlen

Attempts:

3 left

💡 Hint

Common Mistakes

Using a method like length() instead of the property length.

Confusing length with size or count.

The length property returns the number of characters in a string in TypeScript.

2fill in blank

medium

Complete the code to get the character at index 2.

DSA Typescript

const str = "world";
const char = str.[1](2);

Drag options to blanks, or click blank then click option'

AcharCodeAt

Bslice

CcharAt

Dsubstring

Attempts:

3 left

💡 Hint

Common Mistakes

Using charCodeAt which returns a number, not a character.

Using substring or slice which return substrings, not single characters.

The charAt method returns the character at the specified index in a string.

3fill in blank

hard

Fix the error in the code to convert the string to uppercase.

DSA Typescript

const str = "test";
const upper = str.[1]();

Drag options to blanks, or click blank then click option'

Atouppercase

BtoUpperCase

Cuppercase

DtoUpper

Attempts:

3 left

💡 Hint

Common Mistakes

Using incorrect casing like 'touppercase'.

Using incomplete method names like 'toUpper'.

The correct method to convert a string to uppercase in TypeScript is toUpperCase().

4fill in blank

hard

Fill both blanks to create a substring from index 1 to 4.

DSA Typescript

const str = "example";
const sub = str.[1]([2], 4);

Drag options to blanks, or click blank then click option'

Asubstring

B1

C2

Dslice

Attempts:

3 left

💡 Hint

Common Mistakes

Using slice with wrong start index.

Confusing start and end indexes.

The substring method extracts characters from index 1 up to but not including index 4.

5fill in blank

hard

Fill all three blanks to create a new string with the first letter capitalized.

DSA Typescript

const str = "hello";
const capitalized = str.[1](0, 1).[2]() + str.[3](1);

Drag options to blanks, or click blank then click option'

Asubstring

BtoUpperCase

Dslice

Attempts:

3 left

💡 Hint

Common Mistakes

Using slice instead of substring for the first part.

Not converting the first letter to uppercase.

Using substring for the last part instead of slice.

This code takes the first letter using substring(0, 1), converts it to uppercase, then adds the rest of the string using slice(1).

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