0139_word_break.py

from all_import import *

'''
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.
'''

class TrieNode:
    def __init__(self):
        self.isWord = False
        self.child = defaultdict(TrieNode)
    
    def addWord(self, word):
        cur = self
        for c in word:
            cur = cur.child[c]
        cur.isWord = True
        
class Solution:
    def wordBreak(self, s: str, wordDict: List[str]) -> bool:
        root = TrieNode()
        for word in wordDict:
            root.addWord(word)
            
        n = len(s)
        dp = [False] * (n+1)
        dp[n] = True

        for i in range(n-1, -1, -1):
            cur = root
            for j in range(i+1, n+1):
                c = s[j-1]
                if c not in cur.child: 
                    break
                cur = cur.child[c]
                if cur.isWord and dp[j]:
                    dp[i] = True
                    break
        
        return dp[0]