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
```

# Solution

The problem can be divided into sub problem: if the string can be divided into two and each part satisfies the requirement, then the string itself satisfies the requirement.

The DP part of the problem can be like this: the substring up to but not including index `i`

satisfies the question. To solve this, we can further divide the substring (up to `i`

) into two parts, `s[0:j]`

and `s[j:i]`

. `dp[j]`

provides the answer for `s[0:j]`

and we only need to verify `s[j:i]`

is in `wordDict`

or not. If there exists a `j`

that satisfies both requirements, then `dp[i]`

is True.

```
class Solution(object):
def wordBreak(self, s, wordDict):
"""
:type s: str
:type wordDict: List[str]
:rtype: bool
"""
wordDict = set(wordDict)
# a substring [0:i] satisfies the requirement
dp = [False for _ in range(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 wordDict:
dp[i] = True
break
return dp[-1]
```