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
andwordDict[i]
consist of only lowercase English letters.- All the strings of
wordDict
are unique.
Solution
We can apply the backtracking1 template to solve this problem. Fill in the logic.
is_leaf
:start_index == len(s)
, when all the letters are used.get_edges
:w = s[start_index:end_index+1]
wherestart_index <= end_index < len(s)
, are the possible words starting atstart_index
.is_valid
: isw in wordDict
?w
is valid if it's in the dictionary.
Implementation
1def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
2 def dfs(start_index, path):
3 if start_index == len(s):
4 ans.append(" ".join(path))
5 return
6 for end_index in range(start_index, len(s)):
7 w = s[start_index:end_index+1]
8 if w in wordDict:
9 path.append(w)
10 dfs(end_index+1, path)
11 path.pop()
12 ans = []
13 dfs(0, [])
14 return ans
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