How Strings Work Differently Across Languages in DSA Typescript - Debugging & Practice

Challenge - 5 Problems

❓ Predict Output

intermediate

What is the output of this TypeScript string code?

Consider this TypeScript code that manipulates strings. What will it print?

DSA Typescript

const s = 'hello';
const result = s[0] + s[1] + s.length;
console.log(result);

A "he5"

B "hello5"

C "h1"

D "he"

💡 Hint

Remember that s.length gives the number of characters in the string.

The code takes the first two characters 'h' and 'e' from the string 'hello' and adds the length of the string, which is 5. So the output is 'he5'.

🧠 Conceptual

intermediate

How does TypeScript handle string immutability?

Which statement best describes string immutability in TypeScript?

A Strings in TypeScript are mutable only if declared with var.

B Strings in TypeScript are mutable; you can change characters by index.

C Strings in TypeScript are mutable only if declared with let.

D Strings in TypeScript are immutable; you cannot change characters directly.

💡 Hint

Think about whether you can assign a new character to a string index.

Strings in TypeScript (like in JavaScript) are immutable, meaning you cannot change individual characters by index. You must create a new string to change content.

❓ Predict Output

advanced

What is the output of this TypeScript string concatenation?

Analyze the following TypeScript code and determine its output.

DSA Typescript

const a = 'abc';
const b = 'def';
const c = a + b.length + b[0];
console.log(c);

A "abcdef"

B "abc3"

C "abc3d"

D "abcde"

💡 Hint

Remember that b.length is a number and adding it to a string converts it to string.

The code concatenates 'abc' + '3' + 'd' resulting in 'abc3d'.

🔧 Debug

advanced

Which option causes a runtime error in this TypeScript string code?

Given the code snippet below, which option will cause a runtime error when executed?

DSA Typescript

const str: string = 'test';
const char = str.charAt(10);
console.log(char.length);

A Runtime error: undefined has no property 'length'

B No error, prints 0

C Syntax error due to invalid charAt usage

D Runtime error: string index out of range

💡 Hint

Check what charAt returns when index is out of range.

charAt returns an empty string '' if index is out of range, so char.length is 0 and no error occurs.

🚀 Application

expert

How to correctly reverse a string in TypeScript?

Which option correctly reverses the string 'hello' in TypeScript?

A 'hello'.split('').reverse().join('')

B 'hello'.reverse()

C 'hello'.split().reverse().join()

D 'hello'.join('').reverse().split('')

💡 Hint

Strings do not have a reverse method directly.

To reverse a string, split it into an array of characters, reverse the array, then join back into a string.

Practice

1. In TypeScript, what is the type of a string when you use single quotes like 'hello' compared to double quotes like "hello"?

easy

A. Both are of type string and behave the same

B. Single quotes create a character, double quotes create a string

C. Single quotes create a string, double quotes create a character

D. Single quotes are invalid syntax in TypeScript

Solution

1. Step 1: Understand string literals in TypeScript

   TypeScript treats both single and double quotes as string delimiters, so both create string values.

2. Step 2: Confirm no character type in TypeScript

   Unlike some languages, TypeScript does not have a separate character type; single quotes do not create characters.

3. Final Answer:

   Both are of type string and behave the same -> Option A

4. Quick Check:

   Single and double quotes create strings = A [OK]

Hint: Single and double quotes both create strings in TypeScript [OK]

Common Mistakes:

• Thinking single quotes create characters
• Confusing TypeScript with languages like C
• Assuming single quotes are invalid

2. Which of the following is the correct way to declare a multi-line string in TypeScript?

easy

A. Using double quotes with + operator: const s = "line1" + "line2";

B. Using single quotes with \n: const s = 'line1\nline2';

C. Using backticks: const s = `line1\nline2`;

D. Using triple quotes: const s = '''line1\nline2''';

Solution

1. Step 1: Identify multi-line string syntax in TypeScript

   TypeScript supports template literals using backticks (`) which allow multi-line strings directly.

2. Step 2: Check other options for correctness

   Single and double quotes require explicit \n for new lines; triple quotes are invalid in TypeScript.

3. Final Answer:

   Using backticks: const s = `line1\nline2`; -> Option C

4. Quick Check:

   Backticks allow multi-line strings = A [OK]

Hint: Use backticks for multi-line strings in TypeScript [OK]

Common Mistakes:

• Using triple quotes like Python
• Expecting single or double quotes to support multi-line directly
• Forgetting to escape newlines in single/double quotes

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

const s = `Hello\nWorld`;
console.log(s.length);

medium

A. Error

B. 12

C. 10

D. 11

Solution

1. Step 1: Understand the string content

   The string is a template literal with an actual newline character between Hello and World, not the two-character escape sequence.

2. Step 2: Count characters including the newline

   Characters: H(1) e(2) l(3) l(4) o(5) \n is a single newline character (counts as 1), W(6) o(7) r(8) l(9) d(10). So length is 11.

3. Final Answer:

   11 -> Option D

4. Quick Check:

   Template literals interpret \n as newline character = A [OK]

Hint: Template literals interpret \n as a single newline character [OK]

Common Mistakes:

• Assuming \n is two characters in length
• Counting only visible characters
• Expecting runtime error due to escape

4. Identify the error in this TypeScript code snippet:

const greeting = 'Hello;
console.log(greeting);

medium

A. Missing closing single quote for the string

B. Incorrect use of backticks instead of single quotes

C. Missing semicolon after console.log statement

D. No error, code runs fine

Solution

1. Step 1: Check string syntax

   The string starts with a single quote but does not have a closing single quote before the newline.

2. Step 2: Understand string literal rules

   Single-quoted strings cannot span multiple lines without escape; missing closing quote causes syntax error.

3. Final Answer:

   Missing closing single quote for the string -> Option A

4. Quick Check:

   Unclosed string literal = D [OK]

Hint: Strings must be closed before newline unless using backticks [OK]

Common Mistakes:

• Thinking backticks are required here
• Ignoring missing quote errors
• Assuming semicolon omission causes error

5. Given the following TypeScript code, what will be the output?

const s1 = 'café';
const s2 = 'cafe\u0301';
console.log(s1 === s2);

Note: \u0301 is a Unicode combining acute accent.

hard

A. true

B. false

C. SyntaxError

D. RuntimeError

Solution

1. Step 1: Understand Unicode normalization in strings

   s1 uses a single character 'é' (U+00E9), while s2 uses 'e' plus combining accent (U+0301).

2. Step 2: Compare strings with different Unicode forms

   Direct equality compares code units; different Unicode sequences mean s1 !== s2.

3. Final Answer:

   false -> Option B

4. Quick Check:

   Unicode composed vs decomposed forms differ = C [OK]

Hint: Unicode composed and decomposed strings are not equal by default [OK]

Common Mistakes:

• Assuming visually same means equal
• Ignoring Unicode combining characters
• Expecting automatic normalization in equality