Word Break II LeetCode Solution | Easy Approach

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Word Break II Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
Output: ["cats and dog","cat sand dog"]

Example 2:

Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
Output: ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: []

Constraints:

  • 1 <= s.length <= 20
  • 1 <= wordDict.length <= 1000
  • 1 <= wordDict[i].length <= 10
  • s and wordDict[i] consist of only lowercase English letters.
  • All the strings of wordDict are unique.

Word Break II Solutions

Time: O(n)
Space: O(n)

C++

class Solution {
 public:
  vector<string> wordBreak(string s, vector<string>& wordDict) {
    unordered_set<string> wordSet{begin(wordDict), end(wordDict)};
    unordered_map<string, vector<string>> memo;
    return wordBreak(s, wordSet, memo);
  }

 private:
  vector<string> wordBreak(const string& s,
                           const unordered_set<string>& wordSet,
                           unordered_map<string, vector<string>>& memo) {
    if (memo.count(s))
      return memo[s];

    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.count(prefix))
        for (const string& word : wordBreak(suffix, wordSet, memo))
          ans.push_back(prefix + " " + word);
    }

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

    return memo[s] = ans;
  }
};

Java

 

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

  private List<String> wordBreak(final String s, Set<String> wordSet,
                                 Map<String, List<String>> memo) {
    if (memo.containsKey(s))
      return memo.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, memo))
          ans.add(prefix + " " + word);
    }

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

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

Python

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

    @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)

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

      return ans

    return wordBreak(s)

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