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LeetCode: Add and Search Word – Data structure design

Posted on January 19, 2018July 26, 2020 by braindenny

Add and Search Word – Data structure design



Similar Problems:

  • CheatSheet: Leetcode For Code Interview
  • CheatSheet: Common Code Problems & Follow-ups
  • Tag: oodesign, trie

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

For example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

Note:
You may assume that all words are consist of lowercase letters a-z.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.


## https://code.dennyzhang.com/add-and-search-word-data-structure-design
## Basic Ideas: Trie tree. Search with dfs or backtracking
##              TrieNode: children
##     search doesn't necessarily means startswith
##     If add you "word" and "words", then ask you whether "word" exists. You should say True
## Complexity: Time O(n), Space O(n), n how many nodes in the Trie Tree

class TrieNode(object):
    def __init__(self):
        self.children = collections.defaultdict(TrieNode)
        self.is_word = False

class WordDictionary(object):

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.root = TrieNode()

    def addWord(self, word):
        """
        Adds a word into the data structure.
        :type word: str
        :rtype: void
        """
        node = self.root
        for ch in word:
            node = node.children[ch]
        node.is_word = True

    def search(self, word):
        """
        Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter.
        :type word: str
        :rtype: bool
        """
        return self.mysearch(word, self.root)

    def mysearch(self, word, node):
        if len(word) == 0:
            # if node.is_word is False, we get a prefix, instead of a match.
            return node.is_word is True

        ch = word[0]
        if ch != '.':
            if ch not in node.children:
                return False
            return self.mysearch(word[1:], node.children[ch])
        else:
            for child in node.children.values():
                if self.mysearch(word[1:], child) is True:
                    return True
            return False

# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)
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Posted in MediumTagged #trie, oodesign, redo

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