If you desire to ensure that in cases where top is not present, then offsetTop should also be absent, you can define that part of the type as a union like so:
type Top =
{ top: number, offsetTop?: number } |
{ top?: never, offsetTop?: never };
In TypeScript, there isn't a direct way to prevent a property key in an object type, but you can come close by specifying that the key is an optional property whose value is the impossible never type. Since there are no actual values of type never, the most logical thing to do is to exclude the property (or perhaps set it to undefined). So, Top either has a top property of type number and an optional offsetProperty also of type number, or it lacks both a top and an offsetTop property.
Similarly for the left/offsetLeft pair:
type Left =
{ left: number, offsetLeft?: number } |
{ left?: never, offsetLeft?: never };
By combining those with Base, you can create Test:
interface Base {
behavior?: "auto" | "smooth"
}
type Test = Base & Top & Left;
As Test represents the intersection of Base, Top, and Left, a value of type Test must adhere to all three definitions.
Let's put it to the test:
let t: Test;
t = { behavior: "auto", top: 1 }; //valid
t = { top: 1, offsetTop: 2 }; // valid
t = { top: 1, offsetTop: 2, left: 3 }; // valid
t = { top: 1, offsetTop: 2, left: 3, offsetLeft: 4 }; // valid
t = { left: 3 }; // valid
t = { left: 3, offsetLeft: 4 }; // valid
t = { offsetTop: 2 }; // error
t = { offsetLeft: 4 }; // error
t = { top: 1, offsetTop: 2, offsetLeft: 4 }; // error
It seems to be working correctly. The compiler accepts valid values while rejecting invalid ones that contain an offsetTop without a top or an offsetLeft without a left.
Playground link to code