# 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:
        words = set()
        for word in wordDict:
            words.add(word)
        dp = [False]*len(s)
        def checkWord(s,ind):
            print(s)
            for i in range(1,len(s)+1): 
                if dp[i+ind-1]:
                    continue
                elif s[:i] in words:
                    print(i,len(s))
                    dp[i+ind-1] = True
                    if i==len(s) or checkWord(s[i:],ind+i): 
                        return True
                
            return False
        
        return checkWord(s,0)
                        
        