Problem description:

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:

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

Example 2:

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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:

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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.

Solution:

For every word we found in s, we add that to the path. Notice we append additional space to the end if there’re string left in s

The length is from 1 ~ len(s)+1 because we want at least one character in the string.

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class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> List[str]:
        self.dic = set(wordDict)
        self.res = []
        def backtrack(s, path):
            if not s:
                self.res.append(path)
                return
            for i in range(1, len(s)+1):
                if s[:i] in self.dic:
                    tmp = path + s[:i] if not s[i:] else path + s[:i] + ' '
                    backtrack(s[i:], tmp)
        backtrack(s, '')
        return self.res

time complexity: $O()$
space complexity: $O()$
reference:
related problem: