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102 lines
2.8 KiB
Python
102 lines
2.8 KiB
Python
"""
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211. Design Add and Search Words Data Structure
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Difficulty: Medium
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https://leetcode.com/problems/design-add-and-search-words-data-structure/
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──────────────────────────────────────────────────
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Design a data structure that supports adding new words and finding if
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a string matches any previously added string.
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Implement the WordDictionary class:
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• WordDictionary() Initializes the object.
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• void addWord(word) Adds word to the data structure, it can be
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matched later.
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• bool search(word) Returns true if there is any string in the data
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structure that matches word or false otherwise. word may contain dots
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'.' where dots can be matched with any letter.
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Example:
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Input
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["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
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[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
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Output
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[null,null,null,null,false,true,true,true]
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Explanation
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WordDictionary wordDictionary = new WordDictionary();
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wordDictionary.addWord("bad");
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wordDictionary.addWord("dad");
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wordDictionary.addWord("mad");
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wordDictionary.search("pad"); // return False
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wordDictionary.search("bad"); // return True
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wordDictionary.search(".ad"); // return True
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wordDictionary.search("b.."); // return True
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Constraints:
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• 1 <= word.length <= 25
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• word in addWord consists of lowercase English letters.
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• word in search consist of '.' or lowercase English letters.
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• There will be at most 2 dots in word for search queries.
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• At most 10^4 calls will be made to addWord and search.
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"""
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class TrieNode:
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def __init__(self):
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self.children = {}
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self.endOfWord = False
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class WordDictionary:
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def __init__(self):
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self.root = TrieNode()
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def addWord(self, word: str) -> None:
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cur = self.root
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for c in word:
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if c not in cur.children:
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cur.children[c] = TrieNode()
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cur = cur.children[c]
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cur.endOfWord = True
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# Leverage Recursion
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def search(self, word: str) -> bool:
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def dfs(j, root):
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cur = root
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for i in range(j, len(word)):
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c = word[i]
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if c == ".":
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for child in cur.children.values():
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if dfs(i + 1, child):
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return True
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return False
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else:
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if c not in cur.children:
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return False
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cur = cur.children[c]
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return cur.endOfWord
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return dfs(0, self.root)
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# Your WordDictionary object will be instantiated and called as such:
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# obj = WordDictionary()
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# obj.addWord(word)
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# param_3 = obj.search(word)
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