Remove all of the duplicate numbers in an array of numbers [duplicate]

Question

I received this question for practice and the wording confused me, as I see 2 results that it might want.

And either way, I'd like to see both solutions.

For example, if I have an array:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

I'm taking this as wanting the final result as either:

let finalResult = [1, 2, 3, 4, 5, 8, 9, 10];

OR:

let finalResult = [1, 9, 10];

The difference between the two being, one just removes any duplicate numbers and leaves the rest and the second just wants any number that isn't a duplicate.

Either way, I'd like to write two functions that does one of each.

This, given by someone else gives my second solution.

let elems = {},

arr2 = arr.filter(function (e) {
   if (elems[e] === undefined) {
       elems[e] = true;
    return true;
  }
  return false;
});
console.log(arr2);

I'm not sure about a function for the first one (remove all duplicates).

Answer 1

Using Set and Array.from()

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

console.log(Array.from(new Set(arr)));

Alternate using regex

regex explanation here

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let res = arr
  .join(',')
  .replace(/(\b,\w+\b)(?=.*\1)/ig, '')
  .split(',')
  .map(Number);

console.log(res);

Alternate using objects

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];

let obj = arr.reduce((acc, val) => Object.assign(acc, {
  [val]: val
}), {});

console.log(Object.values(obj));

Answer 2

Just use a simple array.filter one-liner:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.indexOf(e) == i).sort(function(a, b){return a - b});
console.log(finalResult);

You could use another filter statement if you wanted the second result:

let arr = [1, 2, 4, 2, 3, 3, 4, 5, 5, 5, 8, 8, 9, 10];
let finalResult = arr.filter((e, i, a) => a.filter(f => f == e).length == 1).sort(function(a, b){return a - b});
console.log(finalResult);