AI solution for TypeScript puzzle on BFE.dev
58. implement Diff<A, B>

Here's one possible solution using mapped types and conditional types:

type DiffKeys<A, B> = {  [K in keyof A | keyof B as Exclude<K, keyof A & keyof B>]:     K extends keyof A ? (K extends keyof B ? never : K) : K extends keyof B ? K : never}[keyof A | keyof B];

Let's break it down:

• We start with a mapped type that iterates over each key in keyof A | keyof B. We use the as clause to rename this type variable to K.
• We check if K is not in the intersection of keyof A and keyof B using the Exclude type. If it's not, we include it in the resulting type.
• For each key K, we use a conditional type to check if it exists in keyof A. If it does, we check if it exists in keyof B. If it does, we return never, since this key should not be included in the result. If it doesn't, we return K, since this key is unique to A.
• If K is not in keyof A, we check if it exists in keyof B. If it does, we return K, since this key is unique to B. If it doesn't, we return never.

Finally, we wrap the whole thing in an indexed access type to get the union of all the keys in the resulting object.

With this code, DiffKeys<{a: 1, b: 2}, {b: 1, c: 2}> returns 'a' | 'c'.