s into words based on spaces.pattern, return False.char_to_word and word_to_char.pattern and the corresponding words in s.
a. If the character is not in char_to_word, check if the word is in word_to_char. If it is, return False.
b. If the character is in char_to_word, check if it maps to a different word. If it does, return False.
c. Update char_to_word and word_to_char with the new character-word pair.True.