Discriminated union narrowing in Typescript - Time & Space Complexity
We want to understand how the time it takes to check types in a discriminated union grows as the number of cases increases.
How does the program's work change when we add more options to the union?
Analyze the time complexity of the following code snippet.
type Shape =
| { kind: 'circle'; radius: number }
| { kind: 'square'; size: number }
| { kind: 'rectangle'; width: number; height: number };
function area(shape: Shape): number {
switch (shape.kind) {
case 'circle':
return Math.PI * shape.radius ** 2;
case 'square':
return shape.size ** 2;
case 'rectangle':
return shape.width * shape.height;
}
}
This code calculates the area of a shape by checking its kind and then using the right formula.
Identify the loops, recursion, array traversals that repeat.
- Primary operation: The switch statement checks the shape's kind against each case.
- How many times: It compares once per call, but the number of cases grows with the union size.
As we add more shape types, the switch must check more cases before finding a match.
| Input Size (number of cases) | Approx. Operations (checks) |
|---|---|
| 3 | Up to 3 checks |
| 10 | Up to 10 checks |
| 100 | Up to 100 checks |
Pattern observation: The number of checks grows linearly with the number of cases.
Time Complexity: O(n)
This means the time to find the right case grows in direct proportion to how many cases there are.
[X] Wrong: "Checking a discriminated union is always constant time because it just looks at one property."
[OK] Correct: While the property access is constant, the switch or if-else must compare against each case, so more cases mean more checks.
Understanding how type checks scale helps you explain your code's efficiency clearly and shows you think about how programs behave as they grow.
"What if we replaced the switch with a map from kind to function? How would the time complexity change?"
Practice
What is the main purpose of a discriminated union in TypeScript?
Solution
Step 1: Understand discriminated union concept
A discriminated union uses a shared property (called discriminant) to identify which type is currently in use.Step 2: Identify the purpose of the discriminant property
This shared property helps TypeScript narrow the type safely and access type-specific properties.Final Answer:
To use a common property to distinguish between different types in a union -> Option DQuick Check:
Discriminated union = common property for type narrowing [OK]
- Thinking discriminated unions require identical properties
- Believing unions can't be narrowed by properties
- Confusing discriminated unions with type intersections
Which of the following is the correct syntax to define a discriminated union type in TypeScript?
type Shape = ?
Solution
Step 1: Check for a shared discriminant property
type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } uses 'kind' as a string literal property to distinguish types, which is correct.Step 2: Verify property types and union structure
type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } correctly defines two object types with 'kind' discriminant and specific properties.Final Answer:
type Shape = { kind: 'circle'; radius: number } | { kind: 'square'; side: number } -> Option CQuick Check:
Discriminated union needs a shared literal property [OK]
- Using & instead of | for union
- Not using a literal type for discriminant
- Using number instead of string literal for discriminant
What will be the output of the following TypeScript code?
type Animal =
| { type: 'dog'; barkVolume: number }
| { type: 'cat'; livesLeft: number };
function describe(animal: Animal) {
if (animal.type === 'dog') {
return `Dog barks at volume ${animal.barkVolume}`;
} else {
return `Cat has ${animal.livesLeft} lives left`;
}
}
console.log(describe({ type: 'cat', livesLeft: 7 }));Solution
Step 1: Analyze the input object and type check
The input has type 'cat' and livesLeft 7, matching the second union member.Step 2: Follow the if condition and return value
Since animal.type is 'cat', the else branch runs, returning "Cat has 7 lives left".Final Answer:
"Cat has 7 lives left" -> Option AQuick Check:
Discriminant 'type' guides correct property access [OK]
- Assuming wrong branch runs
- Expecting compile or runtime errors
- Confusing property names between types
Identify the error in this TypeScript code using discriminated unions:
type Vehicle =
| { kind: 'car'; wheels: number }
| { kind: 'bike'; pedals: number };
function getWheels(vehicle: Vehicle) {
if (vehicle.kind === 'car') {
return vehicle.wheels;
} else {
return vehicle.wheels;
}
}Solution
Step 1: Check property access in each branch
In the else branch, vehicle.kind is 'bike', but code accesses vehicle.wheels which does not exist on bike.Step 2: Understand TypeScript error
This causes a compile-time error because 'wheels' is not a property of the bike type.Final Answer:
Accessing 'wheels' on bike type causes error -> Option BQuick Check:
Access only properties valid for narrowed type [OK]
- Assuming all properties exist on all types
- Ignoring discriminant for narrowing
- Thinking missing return type causes error
Given the following discriminated union, how can you safely add a new shape type { kind: 'triangle'; base: number; height: number } and update the area function to handle it?
type Shape =
| { kind: 'circle'; radius: number }
| { kind: 'square'; side: number };
function area(shape: Shape): number {
switch (shape.kind) {
case 'circle':
return Math.PI * shape.radius ** 2;
case 'square':
return shape.side ** 2;
}
}Solution
Step 1: Extend the discriminated union type
Add { kind: 'triangle'; base: number; height: number } to the Shape union to include the new shape.Step 2: Update the area function switch
Add a new case 'triangle' that calculates area as 0.5 * base * height to handle the new shape safely.Final Answer:
Add triangle to Shape union and add a case 'triangle' in switch returning 0.5 * base * height -> Option AQuick Check:
Update union and switch for new discriminant case [OK]
- Forgetting to update the function for new type
- Using default case instead of explicit case
- Removing switch and losing type safety
Ready to prove it?
Pass the test and this pattern moves unknown
