# 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```
```# TLE solution
class Solution:
def helper(self, s):
if not s: return True
return any(self.helper(s[len(word):]) for word in self.wordDict if s.startswith(word))

def wordBreak(self, s: str, wordDict: List[str]) -> bool:
self.wordDict = wordDict
return self.helper(s)```
```# Basic DP
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
dp = [False] * len(s)

for i in range(len(s)):
for word in wordDict:
if s[:i+1].endswith(word) and (dp[i-len(word)] or i-len(word) == -1):
dp[i] = True

return dp[-1]```
```# Advanced DP
class Solution:
def wordBreak(self, s: str, wordDict: List[str]) -> bool:
dp = [True]
for i in range(1, len(s)+1):
dp += any(dp[j] and s[j:i] in wordDict for j in range(i)),
return dp[-1]```