algorithm_fridays_week06.py

# Required function
def shuffle_class(pupils, pupils_to_move):
  if pupils == None or type(pupils) != list or len(pupils) == 0 or type(pupils_to_move) != int:
    return []
  
  # For this solution's approach, normally, we would just take the absolute value of the pupils to move,
  # but we also take the modulus, just in case, to avoid index out of range edge case, and also avoid round trips
  to_move = abs(pupils_to_move) % len(pupils)

  if pupils_to_move == 0 or to_move == 0:
    return pupils

  back_to_front = True if pupils_to_move > 0 else False
  length = len(pupils)

  # We split, join (swapped) and return on the fly
  if back_to_front:
    return pupils[ (length - to_move) :] + pupils[ 0 : (length - to_move)]
  else:
    return pupils[ to_move : ] + pupils[ 0 : to_move]

if __name__ == '__main__':
  pupils = [8, 5, 3, 7]

  pupils_to_move = 2
  shuffled_pupils = shuffle_class(pupils, pupils_to_move)
  print(shuffled_pupils)
  assert shuffled_pupils == [3, 7, 8, 5]

  pupils_to_move = -1
  shuffled_pupils = shuffle_class(pupils, pupils_to_move)
  print(shuffled_pupils)
  assert shuffled_pupils == [5, 3, 7, 8]

Thanks for your feedback @meekg33k, really appreciate it.

Yes, there are other ways to move the elements around without creating a new array.

I wanted to try something different other than sorting in place, so I used this method because the new array created is returned immediately (ending the function call), and in the scope of the function the pupils list passed in would be released when the function ends/exits. It was my thought that that would be optimal enough for space complexity (and this was done because I was trying to avoid tackling time complexity).

Thanks for considering my solution for the $20 prize. I've indicated interest in the raffle draw via sending a DM on twitter. Thank you again.