most_common_word.py

# Given a string paragraph and a string array of the banned words banned, 
# return the most frequent word that is not banned. It is guaranteed there is
# at least one word that is not banned, and that the answer is unique.

# The words in paragraph are case-insensitive and the answer should be returned in lowercase.

 

# Example 1:

# Input: paragraph = "Bob hit a ball, the hit BALL flew far after it was hit.", banned = ["hit"]
# Output: "ball"
# Explanation: 
# "hit" occurs 3 times, but it is a banned word.
# "ball" occurs twice (and no other word does), so it is the most frequent non-banned word in
# the paragraph. 
# Note that words in the paragraph are not case sensitive,
# that punctuation is ignored (even if adjacent to words, such as "ball,"), 
# and that "hit" isn't the answer even though it occurs more because it is banned.
# Example 2:

# Input: paragraph = "a.", banned = []
# Output: "a"
 

# Constraints:

# 1 <= paragraph.length <= 1000
# paragraph consists of English letters, space ' ', or one of the symbols: "!?',;.".
# 0 <= banned.length <= 100
# 1 <= banned[i].length <= 10
# banned[i] consists of only lowercase English letters.

class Solution:
    def mostCommonWord(self, paragraph: str, banned: list[str]) -> str:
        banned_word = set(word.lower() for word in banned)
        words = re.findall(r'[a-z]+', paragraph.lower()) # type: ignore
        mpp = Counter(words) # type: ignore
        most_common = next(c for (c,i) in mpp.most_common() if c not in banned_word)
        return most_common