Counting array inversions in Javascript

inversion.js

function countInversions(array){
  // Note: this uses a variant of merge sort

  //input handlers
  if (array === undefined) throw new Error("Array must be defined to count inversions");
  if (array.length === 0 || array.length === 1) return 0;
  
  var tally = 0; // count for inversions
  sort(array); // merge sort the array and increment tally when there are crossovers
  return tally;


  function sort(arr) {
    if (arr.length === 1) return arr;
    var right = arr.splice(Math.floor(arr.length/2), arr.length - 1);
    return merge(sort(arr), sort(right));
  }
  function merge(left, right){
    var merged = [];
    var l = 0, r = 0;
    var multiplier = 0;
    while (l < left.length || r < right.length){
      if (l === left.length){
        merged.push(right[r]);
        r++;
      } else if (r === right.length){
        merged.push(left[l]);
        l++;
        tally += multiplier;
      } else if (left[l] < right[r]) {
        merged.push(left[l]);
        tally += multiplier;
        l++;
      } else {
        merged.push(right[r]);
        r++;
        multiplier++;
      } 
    }
    return merged;
  }
}

//tests to run in console
/*
console.assert(countInversions([1, 2, 3]) === 0, "Zero inversions for 1, 2, 3");
console.assert(countInversions([1, 3, 2]) === 1, "One inversion for 1, 3, 2");
console.assert(countInversions([1, 3, 2, 4]) === 1, "One inversion for 1, 3, 2, 4");
console.assert(countInversions([1, 3, 2, 4, 5]) === 1, "One inversions for 1, 3, 2, 4, 5");
console.assert(countInversions([3, 1, 2, 4, 5]) === 2, "Two inversions for 3, 1, 2, 4, 5");
console.assert(countInversions([3, 2, 1, 4, 5]) === 3, "Three inversions for 3, 2, 1, 4, 5");
console.assert(countInversions([3, 2, 1, 4, 5, 6]) === 3, "Three inversions for 3, 2, 1, 4, 5, 6");
console.assert(countInversions([6, 5, 4, 3, 2, 1]) === 15, "Fifteen inversions for 654321");
*/

Good solution but you can simplify it a bit by getting rid of tally variable altogether.
Way to do that is just have:

multiplier += (left.length - l)

instead of multiplier++ on Line 37.

Idea here is that number of inversions = Number of elements to the right of current index (l) of left array. Also, I will advise to rename multiplier with just inversions or num_inversions.

Doesn't work properly, try on this array: [2, 3, 9, 2, 9]
Should be 2, actual result 4.

The two inversions here are (1, 3) (a1 = 3 > 2 = a3) and (2, 3) (a2 = 9 > 2 = a3).

Dear Doesn't Work Properly, for example in this case:[1,3,1,2]
Expected is :2
but the result is:3