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 Trie:
def __init__( self ):
self.is_word = False
self.children = {}
def add_words( self, words ):
for word in words:
curr = self
for ch in word:
if ch not in curr.children:
curr.children[ch] = Trie()
curr = curr.children[ch]
curr.is_word = True
class Solution:
def wordBreak( self, s: str, wordDict: List[str] ) -> bool:
trie = Trie()
trie.add_words( wordDict )
n = len( s )
p = [ False ] + [ True ] * ( n )
for i in range( n ):
if p[i] or s[i] not in trie.children:
continue
curr = trie
for j in range( i, n ):
if s[j] not in curr.children:
break
curr = curr.children[ s[j] ]
if curr.is_word:
p[ j + 1 ] = False
return not p[-1]