String Traversal and Character Access in DSA Javascript - Build from Scratch

String Traversal and Character Access

📖 Scenario: Imagine you are building a simple program that reads a word and checks each letter one by one. This is like reading a word aloud slowly, letter by letter.

🎯 Goal: You will create a program that stores a word, sets up a counter to track position, loops through each letter of the word, and prints each letter on its own line.

📋 What You'll Learn

Create a string variable called word with the value "hello".

Create a variable called index and set it to 0.

Use a while loop that runs while index is less than the length of word.

Inside the loop, access the character at position index in word using word.charAt(index).

Print the character inside the loop.

Increase index by 1 each time the loop runs.

💡 Why This Matters

🌍 Real World

Reading and processing text letter by letter is useful in text editors, spell checkers, and simple games.

💼 Career

Understanding string traversal is a basic skill for programming jobs involving text processing, data parsing, and user input handling.

1

Create the string variable

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

DSA Javascript

# Create a string variable called word with value "hello"
// Your code here

Hint

Use const word = "hello"; to create the string.

2

Set up the index counter

Create a variable called index and set it to 0.

DSA Javascript

const word = "hello";
// Create a variable called index and set it to 0
// Your code here

Hint

Use let index = 0; to start counting from zero.

3

Loop through the string characters

Use a while loop that runs while index is less than word.length. Inside the loop, access the character at position index in word using word.charAt(index). Print the character. Increase index by 1 each time the loop runs.

DSA Javascript

const word = "hello";
let index = 0;
// Use a while loop to go through each character in word
// Inside the loop, print the character at position index
// Increase index by 1 each time
// Your code here

Hint

Use while (index < word.length) to loop. Use word.charAt(index) to get each letter. Print it with console.log. Increase index by 1 inside the loop.

4

Print the final output

Run the program to print each character of word on its own line.

DSA Javascript

const word = "hello";
let index = 0;
while (index < word.length) {
    console.log(word.charAt(index));
    index++;
}
// Run the program to see the output

Hint

Each letter of "hello" should print on its own line.

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