String Traversal and Character Access in DSA Typescript - Practice Problems & Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of character access in a loop

What is the output of the following TypeScript code that traverses a string and prints each character with its index?

DSA Typescript

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

A

0:h
1:e
2:l
3:l
4:o

B

h:0
e:1
l:2
l:3
o:4

C hello

D

0:h
1:e
2:l
3:o
4:l

💡 Hint

Remember that str[i] accesses the character at index i, and i is the index number.

The loop runs from 0 to 4 (length of "hello" - 1). Each iteration prints the index and the character at that index separated by a colon.

❓ Predict Output

intermediate

Output of string traversal with conditional character check

What is the output of this TypeScript code that prints only vowels from the string?

DSA Typescript

const text = "typescript";
for (let i = 0; i < text.length; i++) {
  if ('aeiou'.includes(text[i])) {
    console.log(text[i]);
  }
}

A

a
e
i

B

y
i

C

i
u

D

e
i

💡 Hint

Check which characters in "typescript" are vowels (a, e, i, o, u).

The code checks each character if it is in the string "aeiou". Only 'e' and 'i' appear as vowels in "typescript".

🔧 Debug

advanced

Identify the error in string traversal code

What error does this TypeScript code produce when trying to print each character of a string?

DSA Typescript

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

A undefined printed on last iteration

B SyntaxError: Unexpected token

C TypeError: s is not iterable

D RangeError: Index out of bounds

💡 Hint

Check the loop condition and what happens when i equals s.length.

The loop runs from 0 to s.length inclusive, so on the last iteration s[s.length] is undefined, which prints as 'undefined'.

🧠 Conceptual

advanced

Understanding string immutability in traversal

Why can't you change characters of a string directly by index in TypeScript during traversal?

A TypeScript does not support string indexing

B Strings are immutable; characters cannot be changed by index

C Strings are arrays and can be changed by index

D You must use a special method to change characters by index

💡 Hint

Think about whether strings can be changed after creation.

Strings in TypeScript (and JavaScript) are immutable, meaning you cannot change a character by assigning to an index. You must create a new string instead.

🚀 Application

expert

Find the first repeated character in a string

Given the string "swiss", what is the first character that repeats when traversing from left to right?

DSA Typescript

const word = "swiss";
const seen = new Set<string>();
let repeated = '';
for (let i = 0; i < word.length; i++) {
  if (seen.has(word[i])) {
    repeated = word[i];
    break;
  }
  seen.add(word[i]);
}
console.log(repeated);

A i

B w

C s

D No repeated character

💡 Hint

Track characters seen and find the first one that appears again.

The first repeated character in "swiss" is 's' because it appears at index 0 and repeats at index 3.

Practice

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

easy

A. The character 'e'

B. The character 'l'

C. The substring 'll'

D. An error because strings are not indexable

Solution

1. Step 1: Understand string indexing

   Strings in TypeScript can be accessed by index starting at 0, so index 2 refers to 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 character 'l' -> Option B

4. Quick Check:

   str[2] = 'l' [OK]

Hint: Remember string index starts at 0 [OK]

Common Mistakes:

• Thinking index starts at 1
• Confusing character with substring
• Assuming strings are not indexable

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

easy

A. for (let char of text) { console.log(char); }

B. for (let char in text) { console.log(char); }

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

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

Solution

1. Step 1: Understand looping over strings

   Using for...of loops over characters directly, which is the cleanest way.

2. Step 2: Evaluate the options

   The index-based loop using <= text.length causes an out-of-bounds access. for...in iterates over string indices ('0', '1', etc.) rather than characters. for...of correctly iterates over characters. Starting the index at 1 skips the first character and risks out-of-bounds.

3. Final Answer:

   for (let char of text) { console.log(char); } -> Option A

4. Quick Check:

   Use for...of to access characters directly [OK]

Hint: Use for...of to loop characters directly [OK]

Common Mistakes:

• Using for...in which loops keys, not values
• Starting index at 1 instead of 0
• Confusing index with character in loops

3. What is the output of the following code?

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

medium

A. "bdf"

B. ""

C. "abcde"

D. "ace"

Solution

1. Step 1: Analyze the loop increments and indexing

   The loop starts at 0 and increments by 2 each time, so indexes are 0, 2, 4.

2. Step 2: Collect characters at indexes 0, 2, 4

   Characters are s[0] = 'a', s[2] = 'c', s[4] = 'e'. Concatenated result is "ace".

3. Final Answer:

   "ace" -> Option D

4. Quick Check:

   Indexes 0,2,4 characters combined = "ace" [OK]

Hint: Check loop step size and indexes accessed [OK]

Common Mistakes:

• Assuming loop increments by 1
• Confusing indexes with characters
• Expecting all characters concatenated

4. Identify the error in the following code snippet that tries to print each character of word:

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

medium

A. Missing declaration of variable i

B. Using <= instead of < causes infinite loop

C. Indexing starts at 1, skipping first character and causing undefined at end

D. Strings cannot be accessed by index in TypeScript

Solution

1. Step 1: Check loop index range

   The loop starts at 1 and goes up to word.length (4), so indexes accessed are 1,2,3,4.

2. Step 2: Identify indexing issues

   Index 0 is skipped (first character 't'), and index 4 is out of bounds (undefined). So first character is missed and last access is invalid.

3. Final Answer:

   Indexing starts at 1, skipping first character and causing undefined at end -> Option C

4. Quick Check:

   String indexes start at 0, not 1 [OK]

Hint: Loop indexes for strings start at 0 and go to length-1 [OK]

Common Mistakes:

• Starting loop at 1 instead of 0
• Using <= instead of < in loop condition
• Assuming strings are not indexable

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

hard

A. let rev = '';
for (let i = input.length - 1; i >= 0; i--) {
rev += input[i];
}
console.log(rev);

B. let rev = '';
for (let i = 0; i < input.length; i++) {
rev += input[input.length - i];
}
console.log(rev);

C. let rev = '';
for (let i = input.length; i > 0; i--) {
rev += input[i];
}
console.log(rev);

D. let rev = '';
for (let i = input.length; i >= 0; i--) {
rev += input[i];
}
console.log(rev);

Solution

1. Step 1: Understand correct index range for reverse traversal

   Indexes go from input.length - 1 down to 0 inclusive to cover all characters in reverse.

2. Step 2: Verify the index ranges

   Looping from input.length - 1 down to 0 inclusive correctly captures all characters in reverse. A forward loop using input.length - i first accesses an out-of-bounds index. Starting at input.length immediately accesses invalid index. Decrementing from input.length to 0 also starts with out-of-bounds.

3. Final Answer:

   let rev = ''; for (let i = input.length - 1; i >= 0; i--) { rev += input[i]; } console.log(rev); -> Option A

4. Quick Check:

   Loop from length-1 down to 0 for reverse traversal [OK]

Hint: Start loop at length-1 and go down to 0 [OK]

Common Mistakes:

• Starting loop at length instead of length-1
• Using wrong index inside loop causing undefined
• Loop condition off by one causing missing characters