Simple and concise python bottom-up DP

  • 0

    The code is self-explanatory.
    Note that the Set membership test (s[j:i] in wordDict) in python is O(1).

    class Solution(object):
        def wordBreak(self, s, wordDict):
            :type s: str
            :type wordDict: Set[str]
            :rtype: bool
            dp = [False]*(len(s)+1)
            dp[0] = True
            for i in range(1, len(s)+1):
                for j in range(0, i):
                    if dp[j] and s[j:i] in wordDict:
                        dp[i] = True
            return dp[len(s)]

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