# 139. Word Break (Medium)

https://leetcode.com/problems/word-break/

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

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 = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because `"leetcode"` can be segmented as `"leet code"`.
```

Example 2:

```Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because `"`applepenapple`"` can be segmented as `"`apple pen apple`"`.
Note that you are allowed to reuse a dictionary word.
```

Example 3:

```Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false
```

## Solutions

``````class Solution {

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

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

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

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

// recursively break up the strings, return the possible combinations
return recurse(s, dict);
}

private boolean recurse(String str, Set<String> dict) {
// return tree if reach the end
if (str == null || str.isEmpty()) {
return true;
}

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

// return true if str is previously partitioned.
if (partitionable.contains(str)) {
return true;
}

// 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;
}

boolean result = recurse(str.substring(i), dict);

// str.substring(i) can not be partitioned.
if (!result) {
continue;
}

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 false;
}

// At least 1 partition scheme