I don t know about return type in typescript

Question

TYPESCRIPT 3.4.3

I want to make function like this

exportObjectUnit({ a: 1, b: 2, c: 3, d: 4 }, ['b', 'c'])

OUTPUT { a: 1, d: 4 };

I don't know how to typing this function's return type

export const exportObjectKey = <T, K extends keyof T>(value: T, exports: K[]) => {
  const returnValue = {};

  Object.keys(value)
    .filter(key => {
      if (exports.indexOf(key) !== -1) {
        return false;
      }
      return true;
    })
    .map(key => {
      returnValue[key] = value[key];
    });

  return returnValue as T;
};

If I use this function, return value still have types (With the exception of second string array parameter)

----------EDIT----------------

export const exportObjectKey = <T>(value: T, exports: Array<keyof T>) => {
  const returnValue = {};

  Object.keys(value)
    .filter(key => {
      if (exports.indexOf(key as keyof T) !== -1) {
        return false;
      }
      return true;
    })
    .map(key => {
      returnValue[key] = value[key];
    });

  return returnValue as T;
};

I don't know how to return. Removing seconds parameter array property from first object

-----------EDIT 2----------------

type Omit<T, K extends keyof T> = Pick<T, Exclude<keyof T, K>>;

export const exportObjectKey = <T, K extends keyof T>(value: T, omit: K): Omit<T, K> => {
  delete value[omit];
  return value;
};
export const exportObjectKeys = <T, K extends Array<keyof T>>(value: T, removes: K) =>
  removes.reduce((object, key) => exportObjectKey(object, key), value);

// This is not perfect version

const a = { a: 1, b: 2, c: 3 };
const keyOmitOne = exportObjectKey(a, 'b');
// When I type keyOmitOne.
// Type definition available, It works (a, c)

// ------------------------------------------

// But, when I use exportObjectKeys
const b = { a: 1, b: 2, c: 3 };
const keyOmitArray = exportObjectKey(b, ['b', 'c']);
// I thought type definition works (a available)
// But there is no definition in b value)