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

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

const firstChar = str[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 brackets

Trying to access with curly braces

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

2fill in blank

medium

Complete the code to find the length of the string.

DSA Javascript

const length = str[1];

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

A.length()

B(length)

C[length]

D.length

Attempts:

3 left

💡 Hint

Common Mistakes

Using parentheses after length

Trying to access length as an array element

The length of a string is accessed using the .length property without parentheses.

3fill in blank

hard

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

DSA Javascript

const upper = str[1];

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

A.toUpperCase()

B.touppercase()

C.toupperCase()

D.toUppercase()

Attempts:

3 left

💡 Hint

Common Mistakes

Incorrect capitalization of method

Misspelling the method name

The correct method to convert a string to uppercase is toUpperCase() with exact casing.

4fill in blank

hard

Fill both blanks to extract a substring from index 2 to 5.

DSA Javascript

const sub = str[1]([2], 5);

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

A.substring

B.substr

C2

D3

Attempts:

3 left

💡 Hint

Common Mistakes

Using substr which has different parameters

Wrong start index

The substring method extracts characters from the start index up to but not including the end index. So str.substring(2, 5) extracts characters at indices 2, 3, and 4.

5fill in blank

hard

Fill all three blanks to create a new string by repeating 'Hi' 3 times and converting to lowercase.

DSA Javascript

const newStr = '[1]'.[2]([3]).toLowerCase();

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

AHi

Brepeat

C3

DtoUpperCase

Attempts:

3 left

💡 Hint

Common Mistakes

Using toUpperCase instead of repeat

Wrong number of repetitions

The repeat method repeats a string a given number of times. So 'Hi'.repeat(3) creates 'HiHiHi'. Then toLowerCase() converts it to lowercase.

Practice

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

easy

A. Because strings store a sequence of characters that can be accessed and manipulated individually

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

C. Because strings cannot be changed once created

D. Because strings only hold numbers

Solution

1. Step 1: Understand what a string stores

   A string holds multiple characters in a specific order, like a list of letters.

2. Step 2: Recognize string operations

   We can access, change, or count characters in a string, showing it acts like a data structure.

3. Final Answer:

   Because strings store a sequence of characters that can be accessed and manipulated individually -> Option A

4. Quick Check:

   Strings = sequence of characters [OK]

Hint: Strings hold ordered characters you can access one by one [OK]

Common Mistakes:

• Thinking strings are just plain text without structure
• Believing strings only display text without manipulation
• Confusing strings with numbers or other data types

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

easy

A. str[0]

B. str(0)

C. str.first()

D. str{0}

Solution

1. Step 1: Recall JavaScript string indexing syntax

   JavaScript strings use square brackets with zero-based index to access characters.

2. Step 2: Check each option

   Only str[0] correctly accesses the first character; others are invalid syntax.

3. Final Answer:

   str[0] -> Option A

4. Quick Check:

   Use square brackets for string indexing [OK]

Hint: Use square brackets with index to get string characters [OK]

Common Mistakes:

• Using parentheses instead of brackets
• Trying to call string like a function
• Using curly braces which are invalid here

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

const word = 'hello';
console.log(word[1]);

medium

A. h

B. l

C. o

D. e

Solution

1. Step 1: Identify the string and index used

   The string is 'hello'. Index 1 means the second character because indexing starts at 0.

2. Step 2: Find the character at index 1

   At index 0 is 'h', at index 1 is 'e'. So, output is 'e'.

3. Final Answer:

   e -> Option D

4. Quick Check:

   word[1] = 'e' [OK]

Hint: Remember string indexing starts at zero [OK]

Common Mistakes:

• Confusing index 1 with first character
• Thinking indexing starts at 1
• Mixing up characters in the string

4. Find the error in the following code snippet that tries to print each character of a string:

const text = 'abc';
for(let i = 1; i <= text.length; i++) {
  console.log(text[i]);
}

medium

A. The loop should start from i = 0, not 1

B. The loop condition should be i < text.length, not i <= text.length

C. The loop starts from the wrong index and uses the wrong condition

D. There is no error

Solution

1. Step 1: Check loop start index

   String indexing starts at 0, so loop should start at i = 0 to access first character.

2. Step 2: Check loop end condition

   Since indexing is zero-based, loop should run while i < text.length, not <=, to avoid undefined access.

3. Final Answer:

   The loop starts from the wrong index and uses the wrong condition -> Option C

4. Quick Check:

   Loop from 0 to less than length [OK]

Hint: Loop from 0 to less than string length [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= instead of < in loop condition
• Ignoring zero-based indexing

5. Given the string str = 'abana', which code snippet correctly counts how many times the letter 'a' appears?

hard

A. let count = 0;
for(let char of str) {
if(char == 'a') count += 2;
}
console.log(count);

B. let count = 0;
for(let i = 0; i < str.length; i++) {
if(str[i] === 'a') count++;
}
console.log(count);

C. let count = 0;
for(let i = 1; i <= str.length; i++) {
if(str[i] === 'a') count++;
}
console.log(count);

D. let count = 0;
for(let i = 0; i < str.length; i++) {
if(str[i] = 'a') count++;
}
console.log(count);

Solution

1. Step 1: Check loop indexing and condition

   let count = 0; for(let i = 0; i < str.length; i++) { if(str[i] === 'a') count++; } console.log(count); loops from 0 to less than string length, correctly accessing all characters.

2. Step 2: Check condition and increment

   let count = 0; for(let i = 0; i < str.length; i++) { if(str[i] === 'a') count++; } console.log(count); uses strict equality (===) and increments count by 1 for each 'a'. Others have errors: B adds 2 instead of 1, C uses wrong loop range, D uses assignment (=) instead of comparison (===).

3. Final Answer:

   let count = 0; for(let i = 0; i < str.length; i++) { if(str[i] === 'a') count++; } console.log(count); -> Option B

4. Quick Check:

   Loop 0 to length, if char === 'a' count++ [OK]

Hint: Use correct loop and strict equality to count chars [OK]

Common Mistakes:

• Using assignment (=) instead of comparison (===)
• Looping with wrong start or end indexes
• Incrementing count incorrectly