Question
Formatted question description: https://leetcode.ca/all/211.html
211. Add and Search Word - Data structure design
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.
Example:
addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true
Constraints:
1 <= word.length <= 500
word in addWord consists lower-case English letters.
word in search consist of '.' or lower-case English letters.
At most 50000 calls will be made to addWord and search.
Algorithm
With the structure of the Trie, the only difference is that the search function needs to be rewritten, compared with Trie problem.
Because the . in this question can replace any character, once there is a ., all subtrees need to be searched. As long as one returns true, the entire search function returns true.
Typical DFS problem.
Code
Java
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public class Add_and_Search_Word_Data_structure_design { class WordDictionary { Trie trie; /** Initialize your data structure here. */ public WordDictionary() { trie = new Trie(); } /** Adds a word into the data structure. */ public void addWord(String word) { trie.insert(word); } /** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */ public boolean search(String word) { return trie.search(word); } } /** * Your WordDictionary object will be instantiated and called as such: * WordDictionary obj = new WordDictionary(); * obj.addWord(word); * boolean param_2 = obj.search(word); */ class TrieNode { // R children to node children public TrieNode[] children; private final int R = 26; private boolean isEnd; public TrieNode() { children = new TrieNode[R]; } public boolean containsKey(char ch) { return children[ch -'a'] != null; } public TrieNode get(char ch) { return children[ch -'a']; } public void put(char ch, TrieNode node) { children[ch -'a'] = node; } public void setEnd() { isEnd = true; } public boolean isEnd() { return isEnd; } } class Trie { private TrieNode root; public Trie() { root = new TrieNode(); } // Inserts a word into the trie. public void insert(String word) { TrieNode node = root; // @note:@memorize: iteration is better than recursion during trie building for (int i = 0; i < word.length(); i++) { char currentChar = word.charAt(i); if (!node.containsKey(currentChar)) { node.put(currentChar, new TrieNode()); } node = node.get(currentChar); } node.setEnd(); } // search a prefix or whole key in trie and // returns the node where search ends public boolean search(String word) { return dfs(word, root); } private boolean dfs(String word, TrieNode node) { if (node == null) { return false; } if (word.length() == 0) { return node.isEnd(); } char curLetter = word.charAt(0); if (curLetter == '.') { for (int i = 0; i < 26; i++) { char c = (char)('a' + i); if (dfs(word.substring(1), node.get(c))) { return true; } } } else if (node.containsKey(curLetter)){ return dfs(word.substring(1), node.get(curLetter)); } return false; } } public class WordDictionary_bucket { // bucket: word length -> list of words Map<Integer, List<String>> map = new HashMap<Integer, List<String>>(); // Adds a word into the data structure. public void addWord(String word) { int index = word.length(); if(!map.containsKey(index)){ List<String> list = new ArrayList<String>(); list.add(word); map.put(index, list); }else{ map.get(index).add(word); } } // Returns if the word is in the data structure. A word could // contain the dot character '.' to represent any one letter. public boolean search(String word) { int index = word.length(); if(!map.containsKey(index)){ return false; } List<String> list = map.get(index); if(isWords(word)){ return list.contains(word); } for(String s : list){ if(isSame(s, word)){ return true; } } return false; } boolean isWords(String s){ for(int i = 0; i < s.length(); i++){ // @note: api isLetter() if(!Character.isLetter(s.charAt(i))){ return false; } } return true; } boolean isSame(String a, String search){ if(a.length() != search.length()){ return false; } for(int i = 0; i < a.length(); i++){ if(search.charAt(i) != '.' && search.charAt(i) != a.charAt(i)){ return false; } } return true; } } } -
// OJ: https://leetcode.com/problems/add-and-search-word-data-structure-design/ // Time: // WordDictionary: O(1) // addWord: O(W) // search: O(26^W) // Space: O(NW) struct TrieNode { TrieNode *next[26] = {}; bool end = false; }; class WordDictionary { TrieNode root; bool dfs(TrieNode *node, string &word, int i) { if (!node) return false; if (i == word.size()) return node->end; if (word[i] != '.') return dfs(node->next[word[i] - 'a'], word, i + 1); for (int j = 0; j < 26; ++j) { if (dfs(node->next[j], word, i + 1)) return true; } return false; } public: void addWord(string word) { auto node = &root; for (char c : word) { if (!node->next[c - 'a']) node->next[c - 'a'] = new TrieNode(); node = node->next[c - 'a']; } node->end = true; } bool search(string word) { return dfs(&root, word, 0); } }; -
class TrieNode: def __init__(self): self.neighbours = {} self.isWord = False class Trie: def __init__(self): self.root = TrieNode() def addWord(self, word): root = self.root for i in range(0, len(word)): c = word[i] if c in root.neighbours: root = root.neighbours[c] else: newnode = TrieNode() root.neighbours[c] = newnode root = root.neighbours[c] root.isWord = True class WordDictionary: def __init__(self): self.trie = Trie() self.cache = set([]) def addWord(self, word): self.trie.addWord(word) self.cache.add(word) def search(self, word): if word in self.cache: return True def dfsHelper(root, word, index): if not root: return False if len(word) == index: if root.isWord: return True return False if word[index] != ".": if dfsHelper(root.neighbours.get(word[index], None), word, index + 1): return True else: for nbr in root.neighbours: if dfsHelper(root.neighbours[nbr], word, index + 1): return True return False return dfsHelper(self.trie.root, word, 0)