Comment by _ar7

If you wanted to do O(n) functionally, you could do

    [1, 1, 2, 3, 4].filter(
      (obj => x => {
        if (obj[x]) {
          return false;
        }
        obj[x] = true;
        return true;
      })({}),
    );

What's happening is that you're passing an object (hashmap / hashset) into a function that returns a filtering function, and that object is used inside the closure to track the dupes. It's still a pure function because even though you're mutating the passed in object, the filtering function is still deterministic and referentially transparent.