solution.py

"""
|| HackerRank
    Problem: Making Anagrams
    Level: easy
    Author: amititkgp
    Implementation: apexDev37

    Status: PASSED - '15 test cases passed on HackerRank'
"""

# Copy and paste all your functional code from
# from the HackerRank platform.


def making_anagrams(s1: str, s2: str) -> int:
  if is_anagram(s1, s2):
    return 0

  # Optimization
  truncated = truncate_symmetric_diff(s1, s2)
  deletions = abs(
      len(s1) - len(truncated['s1'])) + abs(len(s2) - len(truncated['s2']))
  if is_anagram(truncated['s1'], truncated['s2']):
    return deletions

  deletions += count_frequency_diff(truncated['s1'], truncated['s2'])
  return deletions


def is_anagram(s1: str, s2: str) -> bool:
  if len(s1) != len(s2):
    return False
  return sorted(s1) == sorted(s2)


def truncate_symmetric_diff(s1: str, s2: str) -> dict[str, str]:
  uncommon_set = set(s1) ^ set(s2)
  uncommon_chars = ''.join(uncommon_set)
  return {'s1': truncate(uncommon_chars, s1),
          's2': truncate(uncommon_chars, s2)}


def truncate(chars: str, given: str) -> str:
  result = ''.join(filter(lambda x: x not in chars, given))
  return result


def count_frequency_diff(s1: str, s2: str) -> int:
  intersection = get_intersection(s1, s2)
  s1_frequency = count_char_frequency(intersection, s1)
  s2_frequency = count_char_frequency(intersection, s2)
  return sum_frequency_diff(s1_frequency, s2_frequency)


def get_intersection(s1: str, s2: str) -> str:
  common_chars: set = set(s1) & set(s2)
  return ''.join(common_chars)


def count_char_frequency(intersection: str, s: str) -> dict[str, int]:
  assert set(intersection) == set(s)
  char_freq = {char: s.count(char) for char in intersection}
  return char_freq


def sum_frequency_diff(s1: dict[str, int], s2: dict[str, int]) -> int:
  assert s1.keys() == s2.keys()
  deletions = sum([abs(s1[char] - s2[char]) for char in s1.keys()])
  return deletions