website/content.en/ChapterFour/0800~0899/0819.Most-Common-Word.md
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.
Example:
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.
Note:
1 <= paragraph.length <= 1000.0 <= banned.length <= 100.1 <= banned[i].length <= 10.paragraph may have uppercase symbols, and even if it is a proper noun.)paragraph only consists of letters, spaces, or the punctuation symbols !?',;.Given a paragraph and a list of banned words. Return the word that appears most frequently and is not in the banned list. The problem guarantees that at least one word is not in the banned list, and that the answer is unique.
Words in the banned list are represented in lowercase and contain no punctuation. Words in the paragraph are case-insensitive. The answer is in lowercase.
package leetcode
import "strings"
func mostCommonWord(paragraph string, banned []string) string {
freqMap, start := make(map[string]int), -1
for i, c := range paragraph {
if c == ' ' || c == '!' || c == '?' || c == '\'' || c == ',' || c == ';' || c == '.' {
if start > -1 {
word := strings.ToLower(paragraph[start:i])
freqMap[word]++
}
start = -1
} else {
if start == -1 {
start = i
}
}
}
if start != -1 {
word := strings.ToLower(paragraph[start:])
freqMap[word]++
}
// Strip the banned words from the freqmap
for _, bannedWord := range banned {
delete(freqMap, bannedWord)
}
// Find most freq word
mostFreqWord, mostFreqCount := "", 0
for word, freq := range freqMap {
if freq > mostFreqCount {
mostFreqWord = word
mostFreqCount = freq
}
}
return mostFreqWord
}