0211: Design Add and Search Words Data Structure
Problem Statement
Design a data structure that supports adding new words and finding if a string matches any previously added string.
Implement the WordDictionary
class:
WordDictionary()
Initializes the object.void addWord(word)
Addsword
to the data structure, it can be matched later.bool search(word)
Returnstrue
if there is any string in the data structure that matchesword
orfalse
otherwise.word
may contain dots'.'
where dots can be matched with any letter.
Example:
Input ["WordDictionary","addWord","addWord","addWord","search","search","search","search"] [[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]] Output [null,null,null,null,false,true,true,true] Explanation WordDictionary wordDictionary = new WordDictionary(); wordDictionary.addWord("bad"); wordDictionary.addWord("dad"); wordDictionary.addWord("mad"); wordDictionary.search("pad"); // return False wordDictionary.search("bad"); // return True wordDictionary.search(".ad"); // return True wordDictionary.search("b.."); // return True
Constraints:
1 <= word.length <= 25
word
inaddWord
consists of lowercase English letters.word
insearch
consist of'.'
or lowercase English letters.- There will be at most
2
dots inword
forsearch
queries. - At most
104
calls will be made toaddWord
andsearch
.
Code Solution
class TrieNode:
def __init__(self, val):
self.val = val
self.end = False
self.children = {}
class WordDictionary:
def __init__(self):
self.root = TrieNode('')
def addWord(self, word):
temp = self.root
for cx, c in enumerate(word):
if c not in temp.children:
temp.children[c] = TrieNode(c)
temp = temp.children[c]
temp.end = True
def search(self, word):
def rec_search(wx, word, init_node):
wlen = len(word)
temp = init_node
for cx in range(wx, wlen):
c = word[cx]
if c == '.':
flag = False
for cc in temp.children:
flag = flag or rec_search(1 + cx, word, temp.children[cc])
return flag
if c not in temp.children:
return False
temp = temp.children[c]
return temp.end
return rec_search(0, word, self.root)