140. Word Break II

Problem Link

Description

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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 10⁵.

Solution

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Python3

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        N = len(s)
        M = len(wordDict)
        res = []
 
        def backtrack(index, curr):
            if index == N:
                res.append(" ".join(curr))
                return
            
            for word in wordDict:
                length = len(word)
                if index + length <= N and s[index : index + length] == word:
                    backtrack(index + length, curr + [word])
 
        backtrack(0, [])
        return res