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 = 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]