JavaScript: collect duplicates in array

duplicates.js

var duplicates = function(array) {
  var iteratee = array.slice(0);
  var seen = [];
  
  return array.filter(function(element) {
    iteratee.shift();
    if (seen.indexOf(element) !== -1) return false;
    
    var hit = iteratee.indexOf(element) !== -1;
    if (hit && seen.indexOf(element) === -1) seen.push(element);
    
    return hit;
  });
}

The idea was to iterate over the array, removing first element from it and then to check if this element is still present in the array.
Then I added a check for already filtered elements.

CoffeeScript:

duplicates = (array) ->
  iteratee = array.slice(0)
  seen = []

  array.filter (element) ->
    iteratee.shift()

    return false if seen.indexOf(element) != -1

    hit = iteratee.indexOf(element) != -1
    seen.push(element) if hit && seen.indexOf(element) == -1

    hit

To use with any array (not recommended):

Array.prototype.duplicates = function() {
  var iteratee = this.slice(0);
  var seen = [];

  return this.filter(function(element) {
    iteratee.shift();
    if (seen.indexOf(element) !== -1) return false;

    var hit = iteratee.indexOf(element) !== -1;
    if (hit && seen.indexOf(element) === -1) seen.push(element);

    return hit;
  });
}

(or)

Array.prototype.duplicates = ->
  iteratee = this.slice(0)
  seen = []

  this.filter (element) ->
    iteratee.shift()

    return false if seen.indexOf(element) != -1

    hit = iteratee.indexOf(element) != -1
    seen.push(element) if hit && seen.indexOf(element) == -1

    hit

Another version based on my uniq.simple Gist

var duplicates = function(array) {
  var iteratee = array.slice(0);

  return array.filter(function(element) {
    iteratee.shift();
    return iteratee.indexOf(element) !== -1;
  }).reduce(function(prev, curr) {
    if (prev.indexOf(curr) === -1) prev.push(curr);
    return prev;
  }, []);
}

(or)

duplicates = (array) ->
  iteratee = array.slice(0)

  array.filter (element) ->
    iteratee.shift()
    iteratee.indexOf(element) != -1
  .reduce (prev, curr) ->
    prev.push(curr) if (prev.indexOf(curr) == -1) 
    return prev
  , []