139. Word Break

Problem Link

Description

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Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

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

 

Example 1:

Input: s = "leetcode", wordDict = ["leet","code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple","pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3:

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

 

Constraints:

• 1 <= s.length <= 300
• 1 <= wordDict.length <= 1000
• 1 <= wordDict[i].length <= 20
• s and wordDict[i] consist of only lowercase English letters.
• All the strings of wordDict are unique.

Solution

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Python3

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        N = len(s)
        mp = set(wordDict)
 
        @cache
        def go(index):
            if index == N:
                return True
            
            curr = ""
            for i in range(index, N):
                curr += s[i]
                if curr in wordDict and go(i + 1):
                    return True
            
            return False
        
        return go(0)