# 139 Word Break

Given a string `s` and a dictionary of strings `wordDict`, return `true` if `s` can be segmented into a space-separated sequence of one or more dictionary words.

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

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
``````

Constraints:

• `1 <= s.length <= 300`
• `1 <= wordDict.length <= 1000`
• `1 <= wordDict[i].length <= 20`
• `s` and `wordDict[i]` consist of only lowercase English letters.
• All the strings of `wordDict` are unique.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 `````` ``````class Solution: def wordBreak(self, s: str, wordDict: List[str]) -> bool: words = set(wordDict) dp = [False] * (len(s) + 1) dp[0] = True for i in range(1, len(s) + 1): for j in range(i): if dp[j] and s[j:i] in words: dp[i] = True break return dp[-1]``````