Implement type Require to set specified Optional properties in an interface as Required

Suggestion

TS interfaces such as Omit, Pick are doing great job helping us re-use current interfaces rather than re-writing them. One interface that I needed today was one to help me set specific property of another as Required. I did my search and it seems such a thing is not implemented. So, I tried and came up with this:

type Require<T, K extends keyof T> = T & { [P in K]: Required<P> };

interface IOptionalProps {
  prop1: string; // required by default
  prop2?: string // optional by default
  prop3?: string // //
}

const foo: Require<IOptionalProps, 'prop2'> = { prop1: 'foo', prop2: 'foo', }; // prop2 can't be left behind
const foobar: Require<IOptionalProps, 'prop2'> = { prop1: 'baz', prop2: 'baz', prop3: 'baz', }; //all props can't be left behind

🔍 Search Terms

• interface require
• interface required
• transform optional required

✅ Viability Checklist

My suggestion meets these guidelines:

• This wouldn't be a breaking change in existing TypeScript/JavaScript code
• This wouldn't change the runtime behavior of existing JavaScript code
• This could be implemented without emitting different JS based on the types of the expressions
• This isn't a runtime feature (e.g. library functionality, non-ECMAScript syntax with JavaScript output, new syntax sugar for JS, etc.)
• This feature would agree with the rest of TypeScript's Design Goals.

⭐ Suggestion

• Review the provided snippet.
• Add it to TS if it goes well with the TS.

📃 Motivating Example

Provided as part of the suggestion.

💻 Use Cases

Provided as part of the suggestion.

Start by reviewing the proposed Require<T, K extends keyof T> definition and its IOptionalProps examples. Determine whether the requested utility type fits TypeScript's existing utility-type design and identify the appropriate compiler tests and documentation; done means the behavior is specified, tested, and documented if accepted.