ReplaceKeys

In this medium-level challenge, you'll implement a ReplaceKeys type that replaces specified keys across union types with new types from a lookup object, setting keys to never when the replacement type is not provided.

type NodeA = {
type: "A"
name: string
flag: number
}
 
type NodeB = {
type: "B"
id: number
flag: number
}
 
type NodeC = {
type: "C"
name: string
flag: number
}
 
type Nodes = NodeA | NodeB | NodeC
 
type ReplacedNodes = ReplaceKeys<
Nodes,
"name" | "flag",
{ name: number; flag: string }
> // {type: 'A', name: number, flag: string} | {type: 'B', id: number, flag: string} | {type: 'C', name: number, flag: string} // would replace name from string to number, replace flag from number to string.
 
type ReplacedNotExistKeys = ReplaceKeys<Nodes, "name", { aa: number }> // {type: 'A', name: never, flag: number} | NodeB | {type: 'C', name: never, flag: number} // would replace name to never

Detailed Explanation

type ReplaceKeys<U, T, Y> = U extends U
  ? {
      [K in keyof U]: K extends T
        ? K extends keyof Y
          ? Y[K]
          : never
        : U[K]
    }
  : never

How it works:

• U extends U is a distributive conditional type pattern that ensures the type distributes over union members. Each member of the union U is processed individually.
• For each union member, we create a mapped type that iterates over its keys K.
• K extends T checks whether the current key is one of the keys we want to replace.
• If the key should be replaced, K extends keyof Y checks whether the replacement lookup object Y actually has a replacement type for this key. If so, we use Y[K] as the new type.
• If the key should be replaced but Y does not contain a matching key, the property type becomes never.
• If the key is not in the replacement set T, we keep the original type U[K].

This challenge helps you understand distributive conditional types and mapped type transformations over union types and how to apply these concepts in real-world scenarios.