# 140. Word Break II (Hard)

https://leetcode.com/problems/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:
[]```

## Solutions

``````class Solution {

// FIXME Be careful that this problem can never return null, if there is no possible combinations, return empty List.

// keep track of the substrings that are unable to partition
Set<String> unpartitionable = new HashSet<>();

// keep track of the possible partition schemes for specific substring
Map<String, List<String>> schemes = new HashMap<>();

public List<String> wordBreak(String s, List<String> wordDict) {
// See if s or wordDict is null
if (s == null || s.length() == 0 || wordDict == null || wordDict.isEmpty()) {
return new ArrayList<>();
}

// transfer words from List to Set
Set<String> dict = new HashSet<>(wordDict);

// recursively break up the strings, return the possible combinations
List<String> ans = recurse(s, dict);
if (ans == null) {
return new ArrayList<>();
}

return ans;
}

private List<String> recurse(String str, Set<String> dict) {
List<String> ans = new ArrayList<>();

// return empty List if reach the end
if (str == null || str.isEmpty()) {
return ans;
}

// return null if str is unpartitionable.
if (unpartitionable.contains(str)) {
return null;
}

// return all the previously partitioned strings if exists
if (schemes.containsKey(str)) {
return schemes.get(str);
}

// keep the count of the possible partition schemes
int count = 0;

// try all the possible splitting schemes, i is the length of current substring.
for (int i = 1; i <= str.length(); i++) {
String sub = str.substring(0, i);
if (!dict.contains(sub)) {
continue;
}

List<String> results = recurse(str.substring(i), dict);

// str.substring(i) can not be partitioned.
if (results == null) {
continue;
}

// create new copy of the ans, in case it tampers the results of substrings of str
results = union(sub, results);

count++;
}

// if no partition scheme available, keep this substring to black list
if (count == 0) {

// must return null, otherwise previous level will take it partitionable.
return null;
}

schemes.put(str, ans);

return ans;
}

private List<String> union(String str, List<String> strs) {
// create a branch new List to hold unioned results
List<String> ans = new ArrayList<>();
if (strs.size() == 0) {

return ans;
}

for (String s : strs) {
ans.add(str + " " + s);
}

return ans;
}
}
``````