<filename>top_400/dp/139_word_break.py
'''
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.
'''

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        dp = [0 for i in range(len(s))]
        for i in range(len(s)):
            for j in range(i):
                if (dp[j] == 1 and s[j+1:i+1] in wordDict):
                    dp[i] = 1
                    break
            if s[:i+1] in wordDict:
                dp[i] = 1
        print(dp)
        return dp[-1] == 1
                    