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# Word Break II LeetCode Solution – Queslers

## Problem – Word Break II

Given a string `s` and a dictionary of strings `wordDict`, add spaces in `s` to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

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: []``````

Constraints:

• `1 <= s.length <= 20`
• `1 <= wordDict.length <= 1000`
• `1 <= wordDict[i].length <= 10`
• `s` and `wordDict[i]` consist of only lowercase English letters.
• All the strings of `wordDict` are unique.

### Word Break II LeetCode Solution in Java

``````public List<String> wordBreak(String s, Set<String> wordDict) {
return DFS(s, wordDict, new HashMap<String, LinkedList<String>>());
}

// DFS function returns an array including all substrings derived from s.
List<String> DFS(String s, Set<String> wordDict, HashMap<String, LinkedList<String>>map) {
if (map.containsKey(s))
return map.get(s);

if (s.length() == 0) {
return res;
}
for (String word : wordDict) {
if (s.startsWith(word)) {
List<String>sublist = DFS(s.substring(word.length()), wordDict, map);
for (String sub : sublist)
res.add(word + (sub.isEmpty() ? "" : " ") + sub);
}
}
map.put(s, res);
return res;
}
``````

### Word Break II LeetCode Solution in Python

``````class Solution(object):
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: Set[str]
:rtype: List[str]
"""
return self.helper(s, wordDict, {})

def helper(self, s, wordDict, memo):
if s in memo: return memo[s]
if not s: return []

res = []
for word in wordDict:
if not s.startswith(word):
continue
if len(word) == len(s):
res.append(word)
else:
resultOfTheRest = self.helper(s[len(word):], wordDict, memo)
for item in resultOfTheRest:
item = word + ' ' + item
res.append(item)
memo[s] = res
return res
``````

### Word Break II LeetCode Solution in C++

``````class Solution {
unordered_map<string, vector<string>> m;

vector<string> combine(string word, vector<string> prev){
for(int i=0;i<prev.size();++i){
prev[i]+=" "+word;
}
return prev;
}

public:
vector<string> wordBreak(string s, unordered_set<string>& dict) {
if(m.count(s)) return m[s]; //take from memory
vector<string> result;
if(dict.count(s)){ //a whole string is a word
result.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,wordBreak(rem,dict));
result.insert(result.end(),prev.begin(), prev.end());
}
}
m[s]=result; //memorize
return result;
}
};
``````
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