TypeScript Recursive Type

Below two code blocks are doing same job and the only difference is below second code block uses variadic tuple. Check the below links for more details.

https://www.typescriptlang.org/docs/handbook/release-notes/typescript-3-7.html#more-recursive-type-aliases

https://www.typescriptlang.org/docs/handbook/release-notes/typescript-4-1.html#recursive-conditional-types

type CompareLength<Left extends any[], Right extends any[]> =
    Left['length'] extends Right['length'] ? 'equal' :
    Left extends [] ? 'shorterLeft' :
    Right extends [] ? 'shorterRight' :
    ((..._: Left) => any) extends ((_: any, ..._1: infer LeftRest) => any) ?
    ((..._: Right) => any) extends ((_: any, ..._1: infer RightRest) => any) ?
    CompareLength<LeftRest, RightRest> :
    never :
    never;

type T1 = CompareLength<[firstName: string], [lastName: string]>;

type T2 = CompareLength<[firstName: string], [lastName: string, age: number]>;

type T3 = CompareLength<[firstName: string, age: number], [lastName: string]>;
type CompareLength<Left extends any[], Right extends any[]> =
    Left['length'] extends Right['length'] ? 'equal' :
    Left extends [] ? 'shorterLeft' :
    Right extends [] ? 'shorterRight' :
    Left extends [any, ...infer L] ? 
    Right extends [any, ...infer R] ? 
    CompareLength<L, R> :
    never :
    never;

type T1 = CompareLength<[firstName: string], [lastName: string]>;

type T2 = CompareLength<[firstName: string], [lastName: string, age: number]>;

type T3 = CompareLength<[firstName: string, age: number], [lastName: string]>;

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