Python Dictionary 29ms beats 90%

  • 0
            if s is None:
                return False
            s_dict = {}
            for letter in s:
                if letter not in s_dict:
                    s_dict[letter] = 1
                    s_dict[letter] += 1
            seen_odd = False
            for value in s_dict.values():
                if value % 2 == 1:
                    if seen_odd == False:
                        seen_odd = True
                        return False
            return True

    What we want to do is see if we have only one odd amount of letters in our word, so we add all letters to a dictionary and iterate through them at the end to see. If we have two odd amounts, we return False

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