Word break

class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        words = set(wordDict)
        q = collections.deque([0])
        seen = set()

        while q:
            start = q.popleft()
            if start == len(s):
                return True
            for end in range(start + 1, len(s) + 1):
                if end in seen:
                    continue
                if s[start:end] in words:
                    q.append(end)
                    seen.add(end)

        return False

Word Break

Difficulty: Medium

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.