140. Word Break II

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. Return all such possible sentences.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.
• You may assume the dictionary does not contain duplicate words.

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:
[]

Approach #1: Recursive. [C++]

class Solution {
private:
    unordered_map<string, vector<string>> m;
    vector<string> combine(string word, vector<string> prev) {
        for (int i = 0; i < prev.size(); ++i) {
            prev[i] += ' ' + word;
        }
        return prev;
    }
    
public:
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        unordered_set<string> wordDict_(wordDict.begin(), wordDict.end());
        if (m.count(s)) return m[s];
        vector<string> result;
        if (wordDict_.find(s) != wordDict_.end()) {
            result.push_back(s);
        }
        for (int i = 1; i < s.size(); ++i) {
            string word = s.substr(i);
            if (wordDict_.find(word) != wordDict_.end()) {
                string rem = s.substr(0, i);
                vector<string> prev = combine(word, wordBreak(rem, wordDict));
                result.insert(result.end(), prev.begin(), prev.end());
            }
        }
        m[s] = result;
        return result;
    }
};