Design Add and Search Words Data Structure

The problem 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) Adds word to the data structure; it can be matched later. bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots ('.'), where dots can be matched with any letter. Examples 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 in addWord consists of lowercase English letters. word in search consists of '.' or lowercase English letters. There will be at most 2 dots in word for search queries. At most 10^4 calls will be made to addWord and search. Brute force solution class WordDictionary { private var store: [String] init() { self.store = [] } func addWord(_ word: String) { self.store.append(word) } func search(_ word: String) -> Bool { let wordArray = Array(word) let wordCount = wordArray.count for w in self.store { let wArray = Array(w) let wCount = wArray.count if wCount != wordCount { continue } var i = 0 while i < wCount { if wArray[i] == wordArray[i] || wordArray[i] == "." { i += 1 } else { break } if i == wCount { return true } } } return false } } Explanation We can solve this problem in a brute-force way by using an array. It is not the most efficient solution because the search operation will take O(m * n) time. We can optimize it by using a Trie (Prefix Tree) data structure. ...