140. Word Break II

Given a string s and a dictionary of strings wordDict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences in any order.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

Example 1:

Input: s = "catsanddog", wordDict = ["cat","cats","and","sand","dog"]
Output: ["cats and dog","cat sand dog"]

Example 2:

Input: s = "pineapplepenapple", wordDict = ["apple","pen","applepen","pine","pineapple"]
Output: ["pine apple pen apple","pineapple pen apple","pine applepen apple"]
Explanation: Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats","dog","sand","and","cat"]
Output: []

Constraints:

• 1 <= s.length <= 20
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 10
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.
• Input is generated in a way that the length of the answer doesn't exceed 10^5.

Code

class Solution(object):
    def wordBreak(self, s, wordDict):
        return self.helper(s, wordDict, {})

    def helper(self, s, wordDict, memo):
        if s in memo: return memo[s]
        if not s: return []

        res = []
        for word in wordDict:
            if not s.startswith(word):
                continue
            if len(word) == len(s):
                res.append(word)
            else:
                resultOfTheRest = self.helper(s[len(word):], wordDict, memo)
                for item in resultOfTheRest:
                    item = word + ' ' + item
                    res.append(item)
        memo[s] = res
        return res