18. 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:
[]

Solution: (Backtracking)

class Solution
{
public:
    vector<string> res;

    void wordUntil(string s, int n, unordered_set<string> &dict, string result)
    {
        if (s == "")
        {
            result.pop_back();
            res.push_back(result);
            return;
        }

        for (int i = 1; i <= n; i++)
        {
            string k = s.substr(0, i);
            if (dict.find(k) != dict.end())
            {
                wordUntil(s.substr(i, n - i), n - i, dict, result + k + " ");
            }
        }
        return;
    }

    vector<string> wordBreak(string s, vector<string> &wordDict)
    {

        unordered_set<string> dict;

        for (int i = 0; i < wordDict.size(); i++)
        {
            dict.insert(wordDict[i]);
        }
        
        int n = s.length();
        string a = "";
        wordUntil(s, n, dict, a);
        return res;
    }
};

Time Complexity: O(n * 2^n)