class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:

        n = len(s)
        dp = [False]*(len(s)+1)
        dp[0] = True

        for i in range(1,n+1):
            for word in wordDict:
                if i-len(word)>=0 and s[i-len(word):i] == word and dp[i-len(word)]:

                    # dp[i] - represents wordBreak ending here
                    # notice dp variable running counter for n+1 variable
                    # cause dp[i] - means word break possible for ith character in string
                    # onlt if dp[i-len(word)] is True and i-len(word):i, notice it is enough
                    # to use i-len(word):i instead of i+1 because then when we are accounting for
                    # for ith iterm in dp variable we essemtially accountting for string ending at i-1
                    # now to account for i-1th character we need to go back from i to i-len(word)

                    dp[i] = dp[i-len(word)]
                    break

        return dp[-1]