ALGORITHM FIDAYS

shuffleAlgo.py

def shuffleClass(pupilsList, noToBeMoved):
  # validate inputs first
  #check if pupils list is a valid list
  if not isinstance(pupilsList, list):
    return []
  #check if noToBeMoved is a valid integer
  if not isinstance(noToBeMoved, int):
    return pupilsList 
  classSize = len(pupilsList)
  #if pupils list contains just one element or noToBeMoved is zero return it
  if classSize <= 1 or noToBeMoved == 0:
    return pupilsList
  #check that the modulos of noToBeMoved and classSize is not zero if it is, return 
  if noToBeMoved >= classSize and abs(noToBeMoved) % classSize == 0:
    return pupilsList
  #no of steps irrespective of direction
  noOfSteps= abs(noToBeMoved) if abs(noToBeMoved) < classSize else abs(noToBeMoved) % classSize
  #set direction for movement
  endToFront = True if noToBeMoved > 0 else False

  if endToFront:#End to Front
    c=[pupilsList.pop() for x in range(noOfSteps)]
    d=[pupilsList.insert(0,x) for x in c]
    return pupilsList
  else:#FrontToEnd
    c=[pupilsList.pop(0) for x in range(noOfSteps)]
    d=[pupilsList.append(x) for x in c]
    return pupilsList
if __name__ == "__main__":
  print(shuffleClass([8,5,3,7], 2)) # return [3,7,8,5]
  print(shuffleClass([4,1,3], -1)) # return [1,3,4]
  print(shuffleClass([8,5,3,7,9,10], -106)) # return [9, 10, 8, 5, 3, 7]
  print(shuffleClass([4,1,3,7,20,90,14], -100)) # return [3, 7, 20, 90, 14, 4, 1]
    

Hello @Sijibomii, thank you participating in Week 6 of #AlgorithmFridays.

Your solution, is absolutely of the most correct solutions that we received. It is robust with really decent edge-case handling, passes the test cases and is readable.

I see that you used a series of pop and insert operations for computing the final result. How expensive do you think those are in terms of time efficiency? Do you think there's a way we could have reduced the number of times, noOfSteps, we had to iteratively pop and insert? Let me know what you think.

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So while your solution was correct, it was not the most optimal solution and sadly, there can be only 1 awardee of the $50 prize.

That said, your solution is being considered for the $20 prize via a raffle draw. The raffle draw will hold today, Friday May 21 at 3.00pm WAT (7.00 am PST)

If you are interested in being a part of the raffle draw, please send an email to uduak@meekg33k.dev or send me a DM on Twitter @meekg33k so I can share the event invite with you.

NB: Only solutions of participants who indicated interest in the raffle draw will be considered.

Thanks once again for participating and see you later today for Week 7 of #AlgorithmFridays.