String Traversal and Character Access in DSA Javascript - Interactive Practice

Practice - 5 Tasks

1fill in blank

easy

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

DSA Javascript

const str = "hello";
const firstChar = str[1];
console.log(firstChar);

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

A.length

B.charAt(1)

C[1]

D[0]

Attempts:

3 left

💡 Hint

Common Mistakes

Using index 1 instead of 0.

Using .length property instead of accessing character.

In JavaScript, you can access the first character of a string using bracket notation with index 0.

2fill in blank

medium

Complete the code to loop through each character of the string using a for loop.

DSA Javascript

const word = "code";
for (let i = 0; i [1] word.length; i++) {
  console.log(word[i]);
}

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

A<=

B<

C>=

D>

Attempts:

3 left

💡 Hint

Common Mistakes

Using '<=' causes index out of range error.

Using '>' or '>=' causes the loop to never run.

The loop should run while i is less than the string length to avoid going out of bounds.

3fill in blank

hard

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

DSA Javascript

const text = "example";
const lastChar = text.charAt(text.length [1] 1);
console.log(lastChar);

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

A-

B+

C*

D/

Attempts:

3 left

💡 Hint

Common Mistakes

Using '+' causes index out of range.

Using '*' or '/' are invalid for index calculation.

To get the last character, subtract 1 from the string length because indices start at 0.

4fill in blank

hard

Fill both blanks to create a loop that prints characters only if they are vowels.

DSA Javascript

const phrase = "hello world";
for (let i = 0; i [1] phrase.length; i++) {
  const ch = phrase[i];
  if ("aeiou".includes(ch) [2] true) {
    console.log(ch);
  }
}

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

A<

B===

C==

D!==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '>=' or '>' in loop condition causes errors.

Using '==' instead of '===' can cause unexpected behavior.

The loop runs while i is less than phrase length. The condition checks if ch is exactly equal to true from includes().

5fill in blank

hard

Fill all three blanks to create an object mapping each character to its uppercase version if it is a letter.

DSA Javascript

const input = "abc123";
const result = {};
for (let i = 0; i [1] input.length; i++) {
  const ch = input[i];
  if (ch [2] "a" && ch [3] "z") {
    result[ch] = ch.toUpperCase();
  }
}
console.log(result);

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

A<

B>=

C<=

D==

Attempts:

3 left

💡 Hint

Common Mistakes

Using '==' instead of comparison operators.

Using '>=' and '<=' in wrong order.

Loop runs while i < input.length. The if condition checks if ch is between 'a' and 'z' inclusive using >= and <=.

Practice

1. What does the expression str[2] return when str = "hello"?

easy

A. The first character of the string, which is 'h'

B. The third character of the string, which is 'l'

C. The length of the string, which is 5

D. An error because strings cannot be accessed by index

Solution

1. Step 1: Understand zero-based indexing in strings

   Strings in JavaScript start counting characters from index 0, so index 2 is the third character.

2. Step 2: Identify the character at index 2 in "hello"

   The characters are: h(0), e(1), l(2), l(3), o(4). So index 2 is 'l'.

3. Final Answer:

   The third character of the string, which is 'l' -> Option B

4. Quick Check:

   str[2] = 'l' [OK]

Hint: Remember string indexes start at 0, so 2 means third character [OK]

Common Mistakes:

• Thinking index 2 is the second character
• Confusing string length with character access
• Believing strings can't be indexed like arrays

2. Which of the following is the correct syntax to loop through each character of a string text in JavaScript?

easy

A. for (let i = 0; i < text.length; i++) { console.log(text[i]); }

B. for (let char of text) { console.log(text[char]); }

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

D. for (let i = 0; i < text.length; i++) { console.log(text.charAt(i+1)); }

Solution

1. Step 1: Check the loop range and indexing

   The loop should start at 0 and go up to less than text.length to cover all characters.

2. Step 2: Verify character access inside the loop

   Accessing characters by text[i] is correct. for (let i = 0; i < text.length; i++) { console.log(text[i]); } uses this correctly.

3. Final Answer:

   for (let i = 0; i < text.length; i++) { console.log(text[i]); } -> Option A

4. Quick Check:

   Loop from 0 to length-1 and access text[i] [OK]

Hint: Loop from 0 to length-1 and use text[i] to access chars [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using text[char] when char is a character, not index
• Accessing index i+1 causing off-by-one errors

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

const s = "world";
let result = "";
for (let i = 0; i < s.length; i++) {
  if (i % 2 === 0) {
    result += s[i];
  }
}
console.log(result);

medium

A. "wrd"

B. "wolrd"

C. "orld"

D. "wrld"

Solution

1. Step 1: Understand the loop and condition

   The loop goes from 0 to 4 (length 5). It adds characters at even indexes (0,2,4) to result.

2. Step 2: Identify characters at even indexes

   Index 0: 'w', 2: 'r', 4: 'd'. So result becomes "wrd".

3. Final Answer:

   "wrd" -> Option A

4. Quick Check:

   Characters at even indexes = "wrd" [OK]

Hint: Pick characters where index % 2 is zero [OK]

Common Mistakes:

• Including characters at odd indexes
• Confusing index positions with characters
• Appending characters incorrectly

4. The following code is intended to print each character of word on a new line, but it prints an extra undefined. What is the problem?

const word = "test";
for (let i = 0; i <= word.length; i++) {
  console.log(word[i]);
}

medium

A. Strings cannot be accessed by index in JavaScript

B. console.log cannot print string characters

C. The variable i is not declared

D. The loop condition should be i < word.length instead of <=

Solution

1. Step 1: Analyze the loop condition

   The loop runs from i = 0 to i = word.length (which is 4). Index 4 is out of bounds for string indices 0-3.

2. Step 2: Identify the fix

   Change the condition to i < word.length to avoid accessing undefined index.

3. Final Answer:

   The loop condition should be i < word.length instead of <= -> Option D

4. Quick Check:

   Loop must stop before length to avoid undefined index [OK]

Hint: Use i < length not <= to avoid out-of-bounds [OK]

Common Mistakes:

• Using <= instead of < in loop condition
• Thinking strings can't be indexed
• Ignoring off-by-one errors

5. Given a string inputStr, which code snippet correctly reverses the string by traversing it from the end to the start?

hard

A. let reversed = '';
for (let i = inputStr.length; i > 0; i--) {
reversed += inputStr[i];
}

B. let reversed = '';
for (let i = 0; i < inputStr.length; i++) {
reversed += inputStr[inputStr.length - i];
}

C. let reversed = '';
for (let i = inputStr.length - 1; i >= 0; i--) {
reversed += inputStr[i];
}

D. let reversed = '';
for (let i = inputStr.length - 1; i > 0; i--) {
reversed += inputStr[i];
}

Solution

1. Step 1: Understand string indexing and loop bounds

   String indices go from 0 to length-1. To reverse, start at length-1 down to 0 inclusive.

2. Step 2: Check each option's loop and indexing

   let reversed = ''; for (let i = inputStr.length - 1; i >= 0; i--) { reversed += inputStr[i]; } loops from length-1 to 0 inclusive and appends characters correctly. Others have off-by-one errors or wrong indices.

3. Final Answer:

   let reversed = ''; for (let i = inputStr.length - 1; i >= 0; i--) { reversed += inputStr[i]; } -> Option C

4. Quick Check:

   Loop backward from last index to 0 and append chars [OK]

Hint: Loop backward from length-1 to 0 and append characters [OK]

Common Mistakes:

• Using wrong loop bounds causing missing first/last char
• Accessing index equal to length (out of bounds)
• Stopping loop before reaching index 0