

def 				A_					(							a   ,				a					):

      """simple docstring"""



      SCREAMING_SNAKE_CASE_			:					List[Any]          =       sorted(a   ,				key=lambda a					: value[2]   ,				reverse=a					)

      # Create a list of size equal to the maximum deadline
      # and initialize it with -1
      SCREAMING_SNAKE_CASE_			:					str          =       max(a   ,				key=lambda a					: value[1]					)[1]
      SCREAMING_SNAKE_CASE_			:					List[Any]          =       [-1] * max_deadline

      # Finding the maximum profit and the count of jobs
      SCREAMING_SNAKE_CASE_			:					Optional[Any]          =       0
      SCREAMING_SNAKE_CASE_			:					List[str]          =       0
      for job in jobs:
            # Find a free time slot for this job
            # (Note that we start from the last possible slot)
            for i in range(job[1] - 1   ,				-1   ,				-1					):
                  if time_slots[i] == -1:
                        SCREAMING_SNAKE_CASE_			:					List[str]          =       job[0]
                        count += 1
                        max_profit += job[2]
                        break
      return [count, max_profit]


if __name__ == "__main__":
    import doctest

    doctest.testmod()









from collections import defaultdict

def 				A_					(							a					):

      """simple docstring"""



      SCREAMING_SNAKE_CASE_			:					defaultdict[str, int]          =       defaultdict(a					)
      # Creating a dictionary containing count of each word
      for word in sentence.split():
            occurrence[word] += 1
      return occurrence


if __name__ == "__main__":
    for word, count in word_occurrence('INPUT STRING').items():
        print(F'{word}: {count}')






