pick function accepting string or array in typescript

Typescript compiler doesn't know what your string will be when you're evaluating it with the split, therefore you must force K[] on it and that will return all the properties of T.

Based on your desired usage, only the second one is viable for getting the desired types.

// i changed the "a" property to a string
const o = { a: 'hello', b: 2, c: 3 };

// <T, K extends keyof T> - here you assign the generics
// T - will be used on "obj" parameter so it can inherit it's properties
// K - will be a property of T
// I typed the "keys" parameter with "string" (to respect your first usage) and K[] (an array of K properties , for the second one)
// At last, we want the function to return K props of T, we have the Pick construct for that.
const pick = <T, K extends keyof T>(obj: T, keys: string | K[]): Pick<T, K> => {
    if (!Array.isArray(keys)) keys = (keys as string).split(',') as K[]; // we know that "keys" is a string, so we'll force the type on it, and we'll force K[] on the .split result, this will return all types from T.
    return keys.reduce((acum, key: K) => (acum[key] = obj[key], acum), {} as T ); // here we mark the accumulator as T, so we know what properties are used.
};

let p1 = pick(o, 'a,c'); // { a: string , b: number, c: number } - You'll get all the keys from obj
let p2 = pick(o, ['a','c']); // { a: string , c: number }

You can have use union

const pick = (obj: Object,  keys: string[] | string) => {
....
}