Description

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences.

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 = "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:
[]

Solutions

1. dfs

# Time: O(?)
# Space: O(n)
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        res = []
        memo = dict()
        return self.dfs(s, res, wordDict, memo)
    
    def dfs(self, s, res, wordDict, memo):
        if s in memo: return memo[s]
        if not s:
            return [""]
        res = []
        for word in wordDict:
            if s[:len(word)] != word: continue
            for r in self.dfs(s[len(word):], res, wordDict, memo):
                res.append(word + ("" if not r else " " + r))
        memo[s] = res
        return res
# 39/39 cases passed (48 ms)
# Your runtime beats 76.51 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (12.8 MB)

References

  1. 140. Word Break II