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Question
Formatted question description: https://leetcode.ca/all/211.html
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
.
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
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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; } } } ############ class Trie { Trie[] children = new Trie[26]; boolean isEnd; } class WordDictionary { private Trie trie; /** Initialize your data structure here. */ public WordDictionary() { trie = new Trie(); } public void addWord(String word) { Trie node = trie; for (char c : word.toCharArray()) { int idx = c - 'a'; if (node.children[idx] == null) { node.children[idx] = new Trie(); } node = node.children[idx]; } node.isEnd = true; } public boolean search(String word) { return search(word, trie); } private boolean search(String word, Trie node) { for (int i = 0; i < word.length(); ++i) { char c = word.charAt(i); int idx = c - 'a'; if (c != '.' && node.children[idx] == null) { return false; } if (c == '.') { for (Trie child : node.children) { if (child != null && search(word.substring(i + 1), child)) { return true; } } return false; } node = node.children[idx]; } return node.isEnd; } } /** * Your WordDictionary object will be instantiated and called as such: * WordDictionary obj = new WordDictionary(); * obj.addWord(word); * boolean param_2 = obj.search(word); */
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// 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); } };
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class Trie: def __init__(self): self.children = [None] * 26 self.is_end = False class WordDictionary: def __init__(self): self.trie = Trie() def addWord(self, word: str) -> None: node = self.trie for c in word: idx = ord(c) - ord('a') if node.children[idx] is None: node.children[idx] = Trie() node = node.children[idx] node.is_end = True def search(self, word: str) -> bool: def search(word, node): for i in range(len(word)): c = word[i] idx = ord(c) - ord('a') if c != '.' and node.children[idx] is None: return False if c == '.': for child in node.children: if child is not None and search(word[i + 1 :], child): return True return False node = node.children[idx] return node.is_end return search(word, self.trie) # Your WordDictionary object will be instantiated and called as such: # obj = WordDictionary() # obj.addWord(word) # param_2 = obj.search(word) ############ 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)
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type WordDictionary struct { root *trie } func Constructor() WordDictionary { return WordDictionary{new(trie)} } func (this *WordDictionary) AddWord(word string) { this.root.insert(word) } func (this *WordDictionary) Search(word string) bool { n := len(word) var dfs func(int, *trie) bool dfs = func(i int, cur *trie) bool { if i == n { return cur.isEnd } c := word[i] if c != '.' { child := cur.children[c-'a'] if child != nil && dfs(i+1, child) { return true } } else { for _, child := range cur.children { if child != nil && dfs(i+1, child) { return true } } } return false } return dfs(0, this.root) } type trie struct { children [26]*trie isEnd bool } func (t *trie) insert(word string) { cur := t for _, c := range word { c -= 'a' if cur.children[c] == nil { cur.children[c] = new(trie) } cur = cur.children[c] } cur.isEnd = true } /** * Your WordDictionary object will be instantiated and called as such: * obj := Constructor(); * obj.AddWord(word); * param_2 := obj.Search(word); */
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using System.Collections.Generic; using System.Linq; class TrieNode { public bool IsEnd { get; set; } public TrieNode[] Children { get; set; } public TrieNode() { Children = new TrieNode[26]; } } public class WordDictionary { private TrieNode root; public WordDictionary() { root = new TrieNode(); } public void AddWord(string word) { var node = root; for (var i = 0; i < word.Length; ++i) { TrieNode nextNode; var index = word[i] - 'a'; nextNode = node.Children[index]; if (nextNode == null) { nextNode = new TrieNode(); node.Children[index] = nextNode; } node = nextNode; } node.IsEnd = true; } public bool Search(string word) { var queue = new Queue<TrieNode>(); queue.Enqueue(root); for (var i = 0; i < word.Length; ++i) { var count = queue.Count; while (count-- > 0) { var node = queue.Dequeue(); if (word[i] == '.') { foreach (var nextNode in node.Children) { if (nextNode != null) { queue.Enqueue(nextNode); } } } else { var nextNode = node.Children[word[i] - 'a']; if (nextNode != null) { queue.Enqueue(nextNode); } } } } return queue.Any(n => n.IsEnd); } }