Given a paragraph and a list of banned words, return the most frequent word that is not in the list of banned words. It is guaranteed there is at least one word that isn’t banned, and that the answer is unique.
Words in the list of banned words are given in lowercase, and free of punctuation. Words in the paragraph are not case sensitive. The answer is in lowercase.
paragraph = “Bob hit a ball, the hit BALL flew far after it was hit.”
banned = [“hit”]
“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.
1 <= paragraph.length <= 1000. 1 <= banned.length <= 100. 1 <= banned[i].length <= 10. The answer is unique, and written in lowercase (even if its occurrences in paragraph may have uppercase symbols, and even if it is a proper noun.) paragraph only consists of letters, spaces, or the punctuation symbols !?',;. There are no hyphens or hyphenated words. Words only consist of letters, never apostrophes or other punctuation symbols.
Time Complexity: O(P + B) where P and B are the max number of words in the paragraph and banned set of words.
Space Complexity: O(P + B)
We’re using the
re module here to split on the words. You can do without it and I don’t think it would be a big deal in an interview to use regex to split on the words; if it was, use
strip to get rid of the punctuation. First we start by getting the set of banned words. This will make it easier to do an
O(1) lookup when we’re looping through the words in the paragraph to make sure a word isn’t a bad word. Finally, we maintain a dictionary of non-banned words and their count in the paragraph. Once we have the count, we can get the max of the dictionary based on value of each key in