Union type syntax and behavior in Typescript - Time & Space Complexity

Time Complexity: Union type syntax and behavior

O(n)

Understanding Time Complexity

We want to understand how the time to check or use union types changes as the input grows.

How does the program's work grow when handling values that can be one of many types?

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: (number | string)[] = [1, 'a', 2, 'b', 3];
inputs.forEach(v => processValue(v));
    

This code processes an array of values that can be numbers or strings, handling each type differently.

Identify Repeating Operations

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

How Execution Grows With Input

Each new item adds one more type check and one more operation based on that type.

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

Pattern observation: The work grows directly with the number of items; doubling items doubles work.

Final Time Complexity

Time Complexity: O(n)

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

Common Mistake

[X] Wrong: "Checking union types adds extra loops or nested checks making it slower than linear."

[OK] Correct: Each value is checked once, so the work grows only with the number of items, not more.

Interview Connect

Understanding how union types affect performance helps you explain your code clearly and reason about efficiency in real projects.

Self-Check

"What if the array contained nested arrays of union types? How would the time complexity change?"

Practice

1. What does a union type in TypeScript allow a variable to do?

easy

A. Hold values of multiple specified types safely

B. Hold only one fixed type value

C. Automatically convert types at runtime

D. Restrict values to a single primitive type

Solution

1. Step 1: Understand union type purpose

   A union type lets a variable accept values from multiple specified types, not just one.

2. Step 2: Compare options with definition

   Only Hold values of multiple specified types safely correctly describes this behavior; others describe fixed or automatic conversions which union types do not do.

3. Final Answer:

   Hold values of multiple specified types safely -> Option A

4. Quick Check:

   Union type = multiple types allowed [OK]

Hint: Union types use | to allow multiple types [OK]

Common Mistakes:

• Thinking union types convert values automatically
• Believing union types restrict to one type only
• Confusing union with intersection types

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

easy

A. let value: string | number;

B. let value: string & number;

C. let value: string, number;

D. let value: string | number[];

Solution

1. Step 1: Recall union type syntax

   Union types use the pipe symbol | between types without parentheses needed.

2. Step 2: Evaluate each option

   let value: string | number; uses correct syntax: string | number. let value: string & number; uses & which means intersection, not union. let value: string, number; uses comma which is invalid. let value: string | number[]; uses number[] which is an array of numbers, not a single number.

3. Final Answer:

   let value: string | number; -> Option A

4. Quick Check:

   Union syntax uses | without commas [OK]

Hint: Use | between types, no commas or & for union [OK]

Common Mistakes:

• Using & instead of | for union types
• Separating types with commas
• Using number[] instead of number

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

function printId(id: string | number) {
  if (typeof id === 'string') {
    console.log('ID is a string: ' + id.toUpperCase());
  } else {
    console.log('ID is a number: ' + id.toFixed(2));
  }
}
printId(42);

medium

A. ID is a string: 42

B. ID is a number: 42

C. ID is a number: 42.00

D. Runtime error due to wrong method

Solution

1. Step 1: Analyze type check and method calls

   The function checks if id is a string or number. For number, it calls toFixed(2) which formats number with 2 decimals.

2. Step 2: Apply input and output

   Input is 42 (number), so else branch runs and prints 'ID is a number: 42.00'.

3. Final Answer:

   ID is a number: 42.00 -> Option C

4. Quick Check:

   Number uses toFixed(2) = 42.00 [OK]

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

Common Mistakes:

• Calling string methods on numbers without check
• Expecting no decimal places in toFixed output
• Ignoring type narrowing with typeof

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

function formatInput(input: string | number) {
  return input.toUpperCase();
}

medium

A. No error, code works fine

B. Error: toUpperCase() not available on number

C. Error: Missing return type annotation

D. Error: Union types cannot be used in functions

Solution

1. Step 1: Check method usage on union type

   Method toUpperCase() exists on string but not on number.

2. Step 2: Understand type safety with union

   Without checking type, calling toUpperCase() on input causes error because input might be number.

3. Final Answer:

   Error: toUpperCase() not available on number -> Option B

4. Quick Check:

   Call methods only after type check [OK]

Hint: Always check type before calling type-specific methods [OK]

Common Mistakes:

• Calling string methods on union without type check
• Assuming union types allow all methods
• Ignoring compiler errors about method availability

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

type Result = string | null;
function process(value: Result) {
  return value?.toUpperCase() ?? 'No value';
}
console.log(process(null));
console.log(process('hello'));

hard

A. No value\nhello

B. null\nhello

C. Error: Cannot call toUpperCase on null

D. No value\nHELLO

Solution

1. Step 1: Understand optional chaining and nullish coalescing

   The expression value?.toUpperCase() calls toUpperCase() only if value is not null or undefined. Otherwise, it returns undefined.

2. Step 2: Apply nullish coalescing fallback

   If the left side is undefined, ?? 'No value' returns 'No value'. So for null, output is 'No value'. For 'hello', it returns 'HELLO'.

3. Final Answer:

   No value HELLO -> Option D

4. Quick Check:

   Optional chaining + ?? handles null safely [OK]

Hint: Use ?. and ?? to safely handle null in unions [OK]

Common Mistakes:

• Expecting error when calling method on null
• Ignoring optional chaining usage
• Confusing null and undefined handling