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[0] = 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]