Problem

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. You may assume the dictionary does not contain duplicate words.

Return all such possible sentences.

For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].

A solution is ["cats and dog", "cat sand dog"].

UPDATE (2017/1/4):
The wordDict parameter had been changed to a list of strings (instead of a set of strings). Please reload the code definition to get the latest changes.

Solution

Time complexity: O(2^n)

Space complexity: O(2^n)

C++

// Author: Huahua
class Solution {
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> dict(wordDict.cbegin(), wordDict.cend());
        return wordBreak(s, dict);
    }
private:
    // >> append({"cats and", "cat sand"}, "dog");
    // {"cats and dog", "cat sand dog"}
    vector<string> append(const vector<string>& prefixes, const string& word) {
        vector<string> results;
        for(const auto& prefix : prefixes)
            results.push_back(prefix + " " + word);
        return results;
    }
    
    const vector<string>& wordBreak(string s, unordered_set<string>& dict) {
        // Already in memory, return directly
        if(mem_.count(s)) return mem_[s];
        
        // Answer for s
        vector<string> ans;
        
        // s in dict, add it to the answer array
        if(dict.count(s)) 
            ans.push_back(s);
        
        for(int j=1;j<s.length();++j) {
            // Check whether right part is a word
            const string& right = s.substr(j);
            if (!dict.count(right)) continue;
            
            // Get the ans for left part
            const string& left = s.substr(0, j);
            const vector<string> left_ans = 
                append(wordBreak(left, dict), right);
            
            // Notice, can not use mem_ here,
            // since we haven't got the ans for s yet.
            ans.insert(ans.end(), left_ans.begin(), left_ans.end());
        }
        
        // memorize and return
        mem_[s].swap(ans);
        return mem_[s];
    }
private:
    unordered_map<string, vector<string>> mem_;
};

Python3

"""
Author: Huahua
Running time: 56 ms
"""
class Solution:
  def wordBreak(self, s, wordDict):
    words = set(wordDict)
    mem = {}
    def wordBreak(s):
      if s in mem: return mem[s]
      ans = []
      if s in words: ans.append(s)
      for i in range(1, len(s)):
        right = s[i:]
        if right not in words: continue        
        ans += [w + " " + right for w in wordBreak(s[0:i])]
      mem[s] = ans
      return mem[s]
    return wordBreak(s)

Related Problems

• [解题报告] Leetcode 139. Word Break 花花酱
• [解题报告] LeetCode 127. Word Ladder
• [解题报告] LeetCode 126. Word Ladder II

The post 花花酱 Leetcode 140. Word Break II appeared first on Huahua's Tech Road.