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819. Most Common Word

Description

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.

Solutions

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• import java.util.regex.Matcher;
  import java.util.regex.Pattern;
  
  class Solution {
      private static Pattern pattern = Pattern.compile("[a-z]+");
  
      public String mostCommonWord(String paragraph, String[] banned) {
          Set<String> bannedWords = new HashSet<>();
          for (String word : banned) {
              bannedWords.add(word);
          }
          Map<String, Integer> counter = new HashMap<>();
          Matcher matcher = pattern.matcher(paragraph.toLowerCase());
          while (matcher.find()) {
              String word = matcher.group();
              if (bannedWords.contains(word)) {
                  continue;
              }
              counter.put(word, counter.getOrDefault(word, 0) + 1);
          }
          int max = Integer.MIN_VALUE;
          String ans = null;
          for (Map.Entry<String, Integer> entry : counter.entrySet()) {
              if (entry.getValue() > max) {
                  max = entry.getValue();
                  ans = entry.getKey();
              }
          }
          return ans;
      }
  }
  

• class Solution {
  public:
      string mostCommonWord(string paragraph, vector<string>& banned) {
          unordered_set<string> s(banned.begin(), banned.end());
          unordered_map<string, int> counter;
          string ans;
          for (int i = 0, mx = 0, n = paragraph.size(); i < n;) {
              if (!isalpha(paragraph[i]) && (++i > 0)) continue;
              int j = i;
              string word;
              while (j < n && isalpha(paragraph[j])) {
                  word.push_back(tolower(paragraph[j]));
                  ++j;
              }
              i = j + 1;
              if (s.count(word)) continue;
              ++counter[word];
              if (counter[word] > mx) {
                  ans = word;
                  mx = counter[word];
              }
          }
          return ans;
      }
  };
  

• class Solution:
      def mostCommonWord(self, paragraph: str, banned: List[str]) -> str:
          s = set(banned)
          p = Counter(re.findall('[a-z]+', paragraph.lower()))
          return next(word for word, _ in p.most_common() if word not in s)
  
  

• func mostCommonWord(paragraph string, banned []string) string {
  	s := make(map[string]bool)
  	for _, w := range banned {
  		s[w] = true
  	}
  	counter := make(map[string]int)
  	var ans string
  	for i, mx, n := 0, 0, len(paragraph); i < n; {
  		if !unicode.IsLetter(rune(paragraph[i])) {
  			i++
  			continue
  		}
  		j := i
  		var word []byte
  		for j < n && unicode.IsLetter(rune(paragraph[j])) {
  			word = append(word, byte(unicode.ToLower(rune(paragraph[j]))))
  			j++
  		}
  		i = j + 1
  		t := string(word)
  		if s[t] {
  			continue
  		}
  		counter[t]++
  		if counter[t] > mx {
  			ans = t
  			mx = counter[t]
  		}
  	}
  	return ans
  }
  

• function mostCommonWord(paragraph: string, banned: string[]): string {
      const s = paragraph.toLocaleLowerCase();
      const map = new Map<string, number>();
      const set = new Set<string>(banned);
      for (const word of s.split(/[^A-z]/)) {
          if (word === '' || set.has(word)) {
              continue;
          }
          map.set(word, (map.get(word) ?? 0) + 1);
      }
      return [...map.entries()].reduce((r, v) => (v[1] > r[1] ? v : r), ['', 0])[0];
  }
  
  

• use std::collections::{ HashMap, HashSet };
  impl Solution {
      pub fn most_common_word(mut paragraph: String, banned: Vec<String>) -> String {
          paragraph.make_ascii_lowercase();
          let banned: HashSet<&str> = banned.iter().map(String::as_str).collect();
          let mut map = HashMap::new();
          for word in paragraph.split(|c| !matches!(c, 'a'..='z')) {
              if word.is_empty() || banned.contains(word) {
                  continue;
              }
              let val = map.get(&word).unwrap_or(&0) + 1;
              map.insert(word, val);
          }
          map.into_iter()
              .max_by_key(|&(_, v)| v)
              .unwrap()
              .0.to_string()
      }
  }
  
  

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