Why union types are needed in Typescript - Performance Analysis

Time Complexity: Why union types are needed

O(n)

Understanding Time Complexity

We want to understand how using union types affects the time it takes for TypeScript code to run.

Specifically, we ask: How does the program's work grow when union types are involved?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

function processValue(value: number | string) {
  if (typeof value === 'number') {
    return value * 2;
  } else {
    return value.toUpperCase();
  }
}

const inputs = [1, 'hello', 3, 'world'];
inputs.forEach(v => console.log(processValue(v)));
    

This code processes an array of numbers and strings, handling each type differently using a union type.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Looping over the array with forEach and type checking each element.
• How many times: Once for each element in the input array.

How Execution Grows With Input

As the input array grows, the number of times we check the type and process values grows directly with it.

Input Size (n)	Approx. Operations
10	About 10 type checks and processing steps
100	About 100 type checks and processing steps
1000	About 1000 type checks and processing steps

Pattern observation: The work grows evenly as the input size grows; doubling input doubles work.

Final Time Complexity

Time Complexity: O(n)

This means the time to run grows in a straight line with the number of items we process.

Common Mistake

[X] Wrong: "Using union types makes the program run slower exponentially because it checks many types."

[OK] Correct: Each item is checked once, so the time grows linearly, not exponentially.

Interview Connect

Understanding how union types affect performance helps you write clear and efficient code, a skill valued in real projects and interviews.

Self-Check

"What if we added nested union types or arrays inside the union? How would the time complexity change?"

Practice

1. Why do we need union types in TypeScript?
let value: string | number;

easy

A. To make variables immutable

B. To allow a variable to hold more than one type of value

C. To force variables to be only one type

D. To automatically convert types at runtime

Solution

1. Step 1: Understand union type declaration

   The syntax string | number means the variable can hold either a string or a number.

2. Step 2: Identify the purpose of union types

   This flexibility allows one variable to store multiple types safely without errors.

3. Final Answer:

   To allow a variable to hold more than one type of value -> Option B

4. Quick Check:

   Union types = multiple allowed types [OK]

Hint: Union types let variables hold multiple types safely [OK]

Common Mistakes:

• Thinking union types make variables immutable
• Believing union types force a single type only
• Confusing union types with automatic type conversion

2. Which of the following is the correct syntax to declare a variable that can be a string or number in TypeScript?

easy

A. let data: string & number;

B. let data: string or number;

C. let data: string | number;

D. let data: string, number;

Solution

1. Step 1: Recall union type syntax

   Union types use the pipe symbol | to separate types.

2. Step 2: Compare options

   let data: string | number; uses | correctly, while others use invalid syntax.

3. Final Answer:

   let data: string | number; -> Option C

4. Quick Check:

   Union syntax uses | symbol [OK]

Hint: Use | to separate types in union declarations [OK]

Common Mistakes:

• Using & instead of | for union types
• Separating types with commas
• Writing 'or' instead of |

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

function format(value: string | number) {
  if (typeof value === 'string') {
    return value.toUpperCase();
  } else {
    return value.toFixed(2);
  }
}

console.log(format('hello'));
console.log(format(3.14159));

medium

A. "HELLO" and 3

B. "hello" and "3.14159"

C. Error because value can be string or number

D. "HELLO" and "3.14"

Solution

1. Step 1: Analyze function behavior for string input

   If input is a string, toUpperCase() converts it to uppercase, so 'hello' becomes 'HELLO'.

2. Step 2: Analyze function behavior for number input

   If input is a number, toFixed(2) formats it to 2 decimal places, so 3.14159 becomes '3.14'.

3. Final Answer:

   "HELLO" and "3.14" -> Option D

4. Quick Check:

   String upper + number fixed decimals = output [OK]

Hint: Check type with typeof before calling type-specific methods [OK]

Common Mistakes:

• Assuming no type check needed and causing errors
• Expecting original number without formatting
• Confusing toFixed output as number instead of string

4. Identify the error in this TypeScript code using union types:

function printLength(value: string | number) {
  console.log(value.length);
}
printLength('test');
printLength(123);

medium

A. Error because number type has no 'length' property

B. Error because string type has no 'length' property

C. No error, code runs fine

D. Error because union types cannot be used in functions

Solution

1. Step 1: Understand property access on union types

   Accessing length on value is unsafe because numbers don't have a length property.

2. Step 2: Identify the error cause

   TypeScript will raise an error to prevent runtime failure when value is a number.

3. Final Answer:

   Error because number type has no 'length' property -> Option A

4. Quick Check:

   Accessing length on number causes error [OK]

Hint: Check type before accessing properties on union types [OK]

Common Mistakes:

• Ignoring type checks and expecting no error
• Thinking string has no length property
• Believing union types disallow functions

5. You want a function that accepts either a string or an Array<string> and returns the total number of characters. Which is the best way to use union types to achieve this?

hard

A. function countChars(input: string | string[]) {
if (typeof input === 'string') {
return input.length;
} else {
return input.join('').length;
}
}

B. function countChars(input: string & string[]) {
return input.length;
}

C. function countChars(input: string | string[]) {
return input.length;
}

D. function countChars(input: any) {
return input.length;
}

Solution

1. Step 1: Understand input types and their length

   A string has a length property; an array of strings also has length, but we want total characters, so we join the array first.

2. Step 2: Check union type handling

   function countChars(input: string | string[]) { if (typeof input === 'string') { return input.length; } else { return input.join('').length; } } uses a type check to handle string and array separately, joining array elements to count total characters correctly.

3. Final Answer:

   function countChars(input: string | string[]) { if (typeof input === 'string') { return input.length; } else { return input.join('').length; } } -> Option A

4. Quick Check:

   Use type check and join array for total chars [OK]

Hint: Use typeof and array methods to handle union types properly [OK]

Common Mistakes:

• Using intersection (&) instead of union (|)
• Not checking type before accessing properties
• Using any type losing type safety