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139. Word Break

Difficulty Topics

Description

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

word-break.py
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        n = len(s)
        dp = [False] * n

        for i in range(n):
            for word in wordDict:
                if s[i - len(word) + 1: i + 1] == word and (dp[i - len(word)] or i - len(word) == -1):
                    dp[i] = True

        return dp[-1]