day-121.cpp

/*
  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 using dynamic programming
class Solution {
  public:
    unordered_map < string, vector < string >> dp;

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

    if (s.empty())
      return {
        ""
      };

    if (dp.find(s) != dp.end())
      return dp[s];

    vector < string > subpart, wholepart;
    for (string word: wordDict) {
      string igot = s.substr(0, word.length());

      if (igot != word) {
        continue;
      } else {
        subpart = help(s.substr(word.length()), wordDict);
      }

      for (string ans: subpart) {
        string space = ans.length() == 0 ? "" : " ";
        wholepart.push_back(word + space + ans);
      }
    }

    return dp[s] = wholepart;
  }

  vector < string > wordBreak(string s, vector < string > & wordDict) {
    dp.clear();
    return help(s, wordDict);
  }
};