Javascript remove all occurrence of duplicate element, leaving the only one that is unique

I want to remove elements that occurr more than once and get the unique element. The array always has 3 elements. Lets say i have an array [2,3,2], then I need to get 3 which is only unique in the array(removing both 2s, because they occur more than once).

I have tried with following code, but surely it doesnot work as expected.

var firstArrTemp = [2,3,2];
var sorted_arr = firstArrTemp.sort();
var unique_element;
for (var i = 0; i < sorted_arr.length - 1; i++) {
    if (sorted_arr[i + 1] != sorted_arr[i]) {
        unique_element=sorted_arr[i];
    }
}

alert(unique_element);

Thanks!

This should do the trick:

Array.prototype.getUnique = function(){
    var uniques = [];
    for(var i = 0, l = this.length; i < l; ++i){
        if(this.lastIndexOf(this[i]) == this.indexOf(this[i])) {
            uniques.push(this[i]);
        }
    }
    return uniques;
}

// Usage:

var a = [2, 6, 7856, 24, 6, 24];
alert(JSON.stringify(a.getUnique()));

console.log(a.getUnique()); // [2, 7856]

To check if a specific item is unique in the array, it just checks if the first index it's found at, matches the last index it's found at.

One alternative with filter() function:

var myArray = [1,2,3,2,2,4,3,7,3].sort();
var uniqueValues = myArray.filter(function(item, i, arr) {
  return (item !== arr[i-1] && item !== arr[i+1]);
});

Where uniqueValues = [1,4,7]

The other answers so far all have O(n log n) time complexity or worse. This can be done in O(n) time though the use of Sets (Set.has has O(1) complexity) instead of nested loops:

// .sort has complexity O(n log n): it's not needed here, avoid it
const getOnlyUniques = (arr) => {
  const foundOnce = new Set();
  const foundTwice = new Set();
  arr.forEach((item) => {
    if (foundOnce.has(item)) {
      foundTwice.add(item);
    }
    foundOnce.add(item);
  });
  return arr.filter(item => !foundTwice.has(item));
};
console.log(getOnlyUniques([2, 3, 2]));