139. Word Break

Medium (Trung bình) Python 🔗 Xem trên LeetCode

📋 Đề Bài

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.

🧠 Thuật Toán & Kỹ Thuật

Dynamic Programming (Quy hoạch động)
⏱️ Thời gian O(n²)
💾 Không gian O(n)

💻 Lời Giải

Python 0139-word-break.py 
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        wordDict = set(wordDict)
        n = len(s)
        dp = [False] * (n + 1)
        dp[n] = True
        
        for i in range(n - 1, -1, -1):
            for j in range(i + 1, n + 1):
                if dp[j] and s[i:j] in wordDict:
                    dp[i] = True
                    break
                    
        return dp[0]