Template literal with unions in Typescript - Time & Space Complexity

Time Complexity: Template literal with unions

O(n * m)

Understanding Time Complexity

We want to understand how the time it takes to create strings using template literals with unions changes as the input size grows.

Specifically, how does the number of possible string combinations affect the work done?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

type Colors = "red" | "green" | "blue";
type Sizes = "small" | "medium" | "large";

type Shirt = `${Colors}-${Sizes}`;

function getAllShirts(): Shirt[] {
  const colors: Colors[] = ["red", "green", "blue"];
  const sizes: Sizes[] = ["small", "medium", "large"];
  const shirts: Shirt[] = [];

  for (const color of colors) {
    for (const size of sizes) {
      shirts.push(`${color}-${size}` as Shirt);
    }
  }
  return shirts;
}
    

This code creates all possible shirt descriptions by combining colors and sizes using template literals with union types.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Nested loops over colors and sizes arrays.
• How many times: Outer loop runs once per color, inner loop runs once per size, total runs = colors x sizes.

How Execution Grows With Input

As the number of colors and sizes grows, the total combinations grow by multiplying these counts.

Input Size (colors x sizes)	Approx. Operations
10 (e.g., 5 colors x 2 sizes)	10 string creations
100 (e.g., 10 colors x 10 sizes)	100 string creations
1000 (e.g., 20 colors x 50 sizes)	1000 string creations

Pattern observation: The work grows proportionally to the product of the input sizes.

Final Time Complexity

Time Complexity: O(n * m)

This means the time grows proportionally to the number of colors times the number of sizes.

Common Mistake

[X] Wrong: "The time grows only with the number of colors or sizes, not both."

[OK] Correct: Because the code combines every color with every size, the total work depends on both inputs multiplied together, not just one.

Interview Connect

Understanding how nested loops multiply work helps you explain performance clearly and shows you can reason about combinations in code.

Self-Check

"What if we added a third union type for styles and nested another loop? How would the time complexity change?"

Practice

1.

Which of the following best describes a template literal with unions in TypeScript?

easy

A. A method to concatenate numbers and strings without type checking.

B. A way to create specific string patterns by combining fixed text and union types.

C. A function that returns a template string without any type restrictions.

D. A way to declare variables without specifying their types.

Solution

1. Step 1: Understand template literals and unions

   Template literals allow embedding expressions in strings. Union types restrict values to a set of options.

2. Step 2: Combine both concepts

   Template literal with unions creates strings with fixed parts and limited variable parts, improving type safety.

3. Final Answer:

   A way to create specific string patterns by combining fixed text and union types. -> Option B

4. Quick Check:

   Template literal + union = specific string patterns [OK]

Hint: Think of fixed text plus limited choices in strings [OK]

Common Mistakes:

• Confusing unions with simple string concatenation
• Ignoring type safety benefits
• Thinking it's only about functions

2.

Which of the following TypeScript declarations correctly uses a template literal with unions?

type Size = "small" | "medium" | "large";
type Message = `Size is ${Size}`;

easy

A. type Message = `Size is ${string}`;

B. type Message = `Size is ${boolean}`;

C. type Message = `Size is ${number}`;

D. type Message = `Size is ${Size}`;

Solution

1. Step 1: Check the union type used inside template literal

   The union type Size is "small" | "medium" | "large" and should be used inside the template literal.

2. Step 2: Match the template literal syntax

   The correct syntax is `Size is ${Size}` to restrict the string to those sizes.

3. Final Answer:

   type Message = `Size is ${Size}`; -> Option D

4. Quick Check:

   Use union type inside template literal exactly [OK]

Hint: Use the union type name inside ${} exactly [OK]

Common Mistakes:

• Replacing union with general types like string or number
• Using incorrect syntax for template literals
• Forgetting backticks around the template

3.

What is the type of result in the following code?

type Color = "red" | "green" | "blue";
type Alert = `Alert: ${Color}`;

const result: Alert = "Alert: green";

medium

A. "Alert: red" | "Alert: green" | "Alert: blue"

B. "Alert: red" | "Alert: green"

C. "Alert: green"

D. string

Solution

1. Step 1: Understand the template literal with union

   The type Alert creates strings by combining "Alert: " with each value in Color.

2. Step 2: Expand the union to all possible strings

   So Alert is the union of "Alert: red", "Alert: green", and "Alert: blue".

3. Final Answer:

   "Alert: red" | "Alert: green" | "Alert: blue" -> Option A

4. Quick Check:

   Template literal expands union to all combined strings [OK]

Hint: Expand union inside template literal to all string combos [OK]

Common Mistakes:

• Assuming only one string value is allowed
• Confusing union with single string
• Using general string type instead of union

4.

Identify the error in this TypeScript code using template literals with unions:

type Status = "success" | "error";
type Message = `Status: ${string}`;

const msg: Message = "Status: success";

medium

A. Using string inside template literal instead of union type Status.

B. Missing backticks around the string value assigned to msg.

C. Incorrect union type declaration for Status.

D. Assigning a string literal to a non-string type.

Solution

1. Step 1: Check the template literal type definition

   The template literal uses ${string} which allows any string, not restricted to Status.

2. Step 2: Identify the correct usage

   It should use ${Status} to restrict values to "success" or "error" only.

3. Final Answer:

   Using string inside template literal instead of union type Status. -> Option A

4. Quick Check:

   Use union type inside template literal for safety [OK]

Hint: Use union type inside ${}, not general string [OK]

Common Mistakes:

• Using string instead of union type inside template
• Ignoring type safety benefits
• Confusing string literals with template literals

5.

Given the following types, which of these assignments is valid without TypeScript errors?

type Direction = "up" | "down";
type Action = "move" | "stop";
type Command = `${Action}-${Direction}`;

const cmd1: Command = "move-up";
const cmd2: Command = "stop-left";
const cmd3: Command = "move-down";
const cmd4: Command = "jump-up";

hard

A. Only cmd2 is valid.

B. All commands are valid.

C. Only cmd1 and cmd3 are valid.

D. Only cmd4 is valid.

Solution

1. Step 1: Understand the Command type

   Command is a template literal combining Action and Direction with a dash, so valid strings are "move-up", "move-down", "stop-up", "stop-down".

2. Step 2: Check each assignment

   cmd1 = "move-up" (valid), cmd2 = "stop-left" (invalid, "left" not in Direction), cmd3 = "move-down" (valid), cmd4 = "jump-up" (invalid, "jump" not in Action).

3. Final Answer:

   Only cmd1 and cmd3 are valid. -> Option C

4. Quick Check:

   Match both parts to union values exactly [OK]

Hint: Check both parts of template literal match union values [OK]

Common Mistakes:

• Assuming any string with dash is valid
• Ignoring union type restrictions
• Confusing Action and Direction values