Why Strings Are a Data Structure Not Just Text in DSA Typescript - Test Your Knowledge

Practice - 5 Tasks

1fill in blank

easy

Complete the code to access the first character of the string.

DSA Typescript

const word = "hello";
const firstChar = word[1];

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

A(0)

B[0]

C{0}

D<0>

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses instead of square brackets

Trying to use curly braces for indexing

Strings in TypeScript can be accessed like arrays using square brackets []. So word[0] gives the first character.

2fill in blank

medium

Complete the code to find the length of the string.

DSA Typescript

const text = "data";
const length = text[1];

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

A.count

B.size

C.length

D.length()

Attempts:

3 left

💡 Hint

Common Mistakes

Using .length() as if it were a function

Using .size which is for other data structures

The length property of a string gives the number of characters. It is accessed without parentheses.

3fill in blank

hard

Fix the error in the code to get the last character of the string.

DSA Typescript

const name = "world";
const lastChar = name[1];

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

A[name.length()]

B[name.length]

C[name.length + 1]

D[name.length - 1]

Attempts:

3 left

💡 Hint

Common Mistakes

Using length as index directly

Adding 1 to length

Using length as a function

Since string indexes start at 0, the last character is at index length - 1.

4fill in blank

hard

Fill both blanks to create a new string with the first and last characters combined.

DSA Typescript

const word = "string";
const newWord = word[1] + word[2];

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

A[0]

B[1]

C[word.length - 1]

D[word.length]

Attempts:

3 left

💡 Hint

Common Mistakes

Using length as last index

Using 1 as first index

The first character is at index 0 and the last character is at index length - 1.

5fill in blank

hard

Fill all three blanks to create an object mapping each character to its index if the character is a vowel.

DSA Typescript

const vowels = ['a', 'e', 'i', 'o', 'u'];
const word = "apple";
const vowelIndices = {};
for (let i = 0; i < word.length; i++) {
  if (vowels[3](word[i])) {
    vowelIndices[[1]] = [2];
  }
};

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

Aword[i]

Bi

C.includes

DindexOf

Attempts:

3 left

💡 Hint

Common Mistakes

Using indexOf instead of includes

Swapping key and value

Using wrong method to check vowels

This creates an object where each vowel character in the word is a key, and its index is the value. The includes method checks if the character is a vowel.

Practice

1. Why are strings considered a data structure and not just plain text?

easy

A. Because strings are only used for displaying text on the screen.

B. Because strings store characters in a specific order and allow operations like indexing and slicing.

C. Because strings cannot be changed once created.

D. Because strings are only sequences of numbers.

Solution

1. Step 1: Understand what a data structure is

   A data structure organizes data to allow efficient access and modification.

2. Step 2: Analyze string properties

   Strings store characters in order and support operations like indexing, slicing, and searching, which are typical data structure features.

3. Final Answer:

   Because strings store characters in a specific order and allow operations like indexing and slicing. -> Option B

4. Quick Check:

   Strings = Ordered character storage [OK]

Hint: Remember: strings hold ordered characters, not just text [OK]

Common Mistakes:

• Thinking strings are just text without structure
• Confusing immutability with data structure properties
• Assuming strings only display data

2. Which of the following is the correct way to access the third character of a string str in TypeScript?

easy

A. str.charAt(3)

B. str(3)

C. str[2]

D. str.get(2)

Solution

1. Step 1: Recall string indexing in TypeScript

   Strings use zero-based indexing, so the first character is at index 0.

2. Step 2: Identify correct syntax for accessing characters

   Using square brackets with index accesses the character at that position, so str[2] gets the third character.

3. Final Answer:

   str[2] -> Option C

4. Quick Check:

   Index 2 = third character [OK]

Hint: Use zero-based index with square brackets to access characters [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Using 1-based indexing
• Using non-existent methods like get()

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

const s = "hello";
console.log(s[1] + s[4]);

medium

A. "eo"

B. "he"

C. "hl"

D. "lo"

Solution

1. Step 1: Identify characters at given indices

   Index 1 of "hello" is 'e', index 4 is 'o'.

2. Step 2: Concatenate characters

   Concatenating 'e' + 'o' results in "eo".

3. Final Answer:

   "eo" -> Option A

4. Quick Check:

   s[1] + s[4] = "e" + "o" = "eo" [OK]

Hint: Remember string indices start at 0; add characters at given indices [OK]

Common Mistakes:

• Using 1-based indexing
• Mixing up indices and characters
• Assuming output is an array

4. Identify the error in this TypeScript code that tries to change the first character of a string:

let word = "hello";
word[0] = "H";
console.log(word);

medium

A. Strings are immutable; you cannot change characters by index.

B. The syntax for changing characters is incorrect; use parentheses instead of brackets.

C. You must declare the string as a constant to modify it.

D. The console.log statement is missing parentheses.

Solution

1. Step 1: Understand string immutability in TypeScript

   Strings cannot be changed after creation; individual characters cannot be assigned new values.

2. Step 2: Analyze the code behavior

   Trying to assign to word[0] does not change the string; no error but no effect.

3. Final Answer:

   Strings are immutable; you cannot change characters by index. -> Option A

4. Quick Check:

   Strings cannot be changed by index assignment [OK]

Hint: Remember strings are immutable; use new string to change content [OK]

Common Mistakes:

• Trying to assign characters directly by index
• Confusing arrays with strings
• Expecting console.log syntax error

5. Given the string str = "data", which TypeScript code correctly creates a new string with the first character capitalized, demonstrating string as a data structure?

hard

A. const newStr = str.replace(0, str[0].toUpperCase());

B. str[0] = str[0].toUpperCase(); const newStr = str;

C. const newStr = str.toUpperCase()[0] + str[1];

D. const newStr = str[0].toUpperCase() + str.slice(1);

Solution

1. Step 1: Understand string immutability and methods

   You cannot change characters by index, but you can create new strings using methods like toUpperCase() and slice().

2. Step 2: Analyze each option

   const newStr = str[0].toUpperCase() + str.slice(1); correctly creates a new string by capitalizing the first character and appending the rest. str[0] = str[0].toUpperCase(); const newStr = str; tries to assign to index (invalid). const newStr = str.toUpperCase()[0] + str[1]; only capitalizes first char but appends only one character. const newStr = str.replace(0, str[0].toUpperCase()); uses replace incorrectly (replace expects substring, not index).

3. Final Answer:

   const newStr = str[0].toUpperCase() + str.slice(1); -> Option D

4. Quick Check:

   Create new string with methods, not index assignment [OK]

Hint: Use string methods to build new strings; never assign by index [OK]

Common Mistakes:

• Trying to assign to string index
• Using replace with index instead of substring
• Not including full string after first char