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.
- The same word in the dictionary may be reused multiple times in the segmentation.
- You may assume the dictionary does not contain duplicate words.
Input: s = "leetcode", wordDict = ["leet", "code"] Output: true Explanation: Return true because "leetcode" can be segmented as "leet code".
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.
Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] Output: false
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,
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 = 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]