"""
Algorithm Type: Divide & Conquer
Time Complexity: O(n*log(n))
"""
numbers = [8, 2, 1, 5, 7, 3, 9, 2, 0, 1]


def count_split_inv(arr, left, right):
  split_inv = ridx = lidx = 0
  size = len(arr)
  lsize = len(left)
  rsize = len(right)
  for i in range(size):
    if lidx != lsize and ridx != rsize:
      if right[ridx] < left[lidx]:
        arr[i] = right[ridx]
        ridx += 1
        split_inv += lsize - lidx
      else:
        arr[i] = left[lidx]
        lidx += 1
    elif lidx == lsize:
      arr[i] = right[ridx]
      ridx += 1
    elif ridx == rsize:
      arr[i] = left[lidx]
      lidx += 1
  return split_inv


def count_inversions(arr) -> int:
  """
  Sort arr and return the number of inversions required.

  >>> numbers
  [8, 2, 1, 5, 7, 3, 9, 2, 0, 1]
  >>> count_inversions(numbers)
  28
  >>> numbers  # numbers has been sorted!!
  [0, 1, 1, 2, 2, 3, 5, 7, 8, 9]
  >>> count_inversions(numbers)
  0
  """
  size = len(arr)
  if size == 1:
    return 0
  mid = int(size/2)
  left = arr[:mid]
  right = arr[mid:]
  l_inv = count_inversions(left)
  r_inv = count_inversions(right)
  split_inv = count_split_inv(arr, left, right)

  return l_inv + r_inv + split_inv

if __name__ == "__main__":
  print(count_inversions(numbers))