algorithm_fridays_week07.py

from datetime import timedelta

def convert_to_minutes(time):
  hours, minutes = time.split(':')
  return timedelta(hours=int(hours), minutes=int(minutes)).seconds // 60

def get_min_diff(times):
  min_diff = None

  for idx, _ in enumerate(times):
    # Do nothing on first iteration to avoid out of range error
    if idx == 0:
      continue

    # Compare adjacent differences
    diff = times[idx] - times[idx - 1]
    if min_diff == None or diff < min_diff:
      min_diff = diff

  return min_diff

def min_time_diff(times):
  """Main function

  ASSUMPTIONS MADE: When a valid input type (a list) is passed,
  each time string in the list is always a valid time format 'HH:MM'

  Seriously though, why would Esther wake up and write down "io:90" or "hello" or "4",
  doesn't make sense, she should probably get more sleep then :).
  """

  # Handle invalid inputs
  if times == None or type(times) != list or len(times) == 0: return []

  # Handle these two cases in O(1)
  if len(times) == 1: return convert_to_minutes(times[0])
  if len(times) == 2: return abs(convert_to_minutes(times[0]) - convert_to_minutes(times[1]))

  # Time complexity is O(n), since it loops through each element once
  # Conversion happens in place, so space complexity remains O(1)
  for idx, time in enumerate(times):
    times[idx] = convert_to_minutes(time)

  # Python built in sort function takes O(n log n)
  times.sort()

  # This takes O(n), since it loops through each element once
  min_diff = get_min_diff(times)

  # Total time complexity would be n + (n log n) + n,
  # which is ultimately (n log n), being a superset of n
  return min_diff

if __name__ == '__main__':
  times = ['16:15', '16:00', '12:20']
  diff = min_time_diff(times)
  print(diff)
  assert diff == 15

Hello @toritsejuFO, congratulations 🎉 your solution has been selected as one of the winning solutions in Week 7 of #AlgorithmFridays.

Your solution is clean and readable and passed the test cases. I particularly like how you structured your code into separate functions. Really neat!

Out of the many winning solutions, only 3 will be selected for the $20 prize in a raffle draw. The raffle draw will hold today, Friday May 28 at 3.00pm WAT (7.00 am PST)

If you are interested in being a part of the raffle draw, please 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 8 of #AlgorithmFridays.

Thanks @meekg33k, appreciate your response.