Word Break II

 

                                          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:
[]

class Solution {
public:
    unordered_map<string,vector<string> > dp;
    
    vector<string> wordBreak(string s, vector<string>& wordDict) {
        if(dp.find(s)!=dp.end())return dp[s];
        vector<string> ans;        
        for(auto st:wordDict)
        {
            if(s.substr(0,st.size())==st)
            {
                if(st.size()==s.size())
                {
                 ans.push_back(st);
                }else{ 
                vector<string> tmp =wordBreak(s.substr(st.size(),s.size()-st.size()),wordDict);
                    for(auto t:tmp)
                        ans.push_back(st+" "+t);
                }
            }
        }
        dp[s]=ans;
         return ans;
    }
    
};