Word Break II

Desicription

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:

1
2
3
4
5
6
7
8
Input:
s = "catsanddog"
wordDict = ["cat", "cats", "and", "sand", "dog"]
Output:
[
"cats and dog",
"cat sand dog"
]

Example 2:

1
2
3
4
5
6
7
8
9
10
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:

1
2
3
4
5
Input:
s = "catsandog"
wordDict = ["cats", "dog", "sand", "and", "cat"]
Output:
[]

Solution

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
class Solution {
private:
unordered_map<string, vector<string>> mp;
unordered_set<string> dict;
vector<string> combine(string s, vector<string> vec) {
for(int i = 0; i < vec.size(); i++)
vec[i] += " " + s;
return vec;
}
vector<string> dfs(string s) {
if(mp.count(s))
return mp[s];
vector<string> res;
if(dict.count(s))
res.push_back(s);
for(int i = 1; i < s.size(); i++) {
string word = s.substr(i);
if(dict.count(word)) {
string rem = s.substr(0, i);
vector<string> prev = combine(word, dfs(rem));
res.insert(res.end(), prev.begin(), prev.end());
}
}
mp[s] = res;
return res;
}
public:
vector<string> wordBreak(string s, vector<string>& wordDict) {
dict = unordered_set<string>(wordDict.begin(), wordDict.end());
return dfs(s);
}
};