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140. Word Break II 👍

  • Time: $O(2^n)$
  • Space: $O(2^n)$
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class Solution {
 public:
  vector<string> wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> wordSet{wordDict.begin(), wordDict.end()};
    unordered_map<string, vector<string>> mem;
    return wordBreak(s, wordSet, mem);
  }

 private:
  vector<string> wordBreak(const string& s,
                           const unordered_set<string>& wordSet,
                           unordered_map<string, vector<string>>& mem) {
    if (const auto it = mem.find(s); it != mem.cend())
      return it->second;

    vector<string> ans;

    // 1 <= prefix.length() < s.length()
    for (int i = 1; i < s.length(); ++i) {
      const string& prefix = s.substr(0, i);
      const string& suffix = s.substr(i);
      if (wordSet.contains(prefix))
        for (const string& word : wordBreak(suffix, wordSet, mem))
          ans.push_back(prefix + " " + word);
    }

    // `wordSet` contains the whole string s, so don't add any space.
    if (wordSet.contains(s))
      ans.push_back(s);

    return mem[s] = ans;
  }
};

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class Solution {
  public List<String> wordBreak(String s, List<String> wordDict) {
    Set<String> wordSet = new HashSet<>(wordDict);
    Map<String, List<String>> mem = new HashMap<>();
    return wordBreak(s, wordSet, mem);
  }

  private List<String> wordBreak(final String s, Set<String> wordSet,
                                 Map<String, List<String>> mem) {
    if (mem.containsKey(s))
      return mem.get(s);

    List<String> ans = new ArrayList<>();

    // 1 <= prefix.length() < s.length()
    for (int i = 1; i < s.length(); ++i) {
      final String prefix = s.substring(0, i);
      final String suffix = s.substring(i);
      if (wordSet.contains(prefix))
        for (final String word : wordBreak(suffix, wordSet, mem))
          ans.add(prefix + " " + word);
    }

    // `wordSet` contains the whole string s, so don't add any space.
    if (wordSet.contains(s))
      ans.add(s);

    mem.put(s, ans);
    return ans;
  }
}

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class Solution:
  def wordBreak(self, s: str, wordDict: list[str]) -> list[str]:
    wordSet = set(wordDict)

    @functools.lru_cache(None)
    def wordBreak(s: str) -> list[str]:
      ans = []

      # 1 <= len(prefix) < len(s)
      for i in range(1, len(s)):
        prefix = s[0:i]
        suffix = s[i:]
        if prefix in wordSet:
          for word in wordBreak(suffix):
            ans.append(prefix + ' ' + word)

      # `wordSet` contains the whole string s, so don't add any space.
      if s in wordSet:
        ans.append(s)

      return ans

    return wordBreak(s)