211

211. Design Add and Search Words Data Structure (Medium)

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.

 

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 <= 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.

Companies:
Facebook, Amazon, Google, Oracle, Apple, Microsoft

Related Topics:
String, Depth-First Search, Design, Trie

Similar Questions:

• Implement Trie (Prefix Tree) (Medium)
• Prefix and Suffix Search (Hard)

Solution 1. Trie

// OJ: https://leetcode.com/problems/add-and-search-word-data-structure-design/
// Author: github.com/lzl124631x
// 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);
    }
};

Solution 2.

// OJ: https://leetcode.com/problems/add-and-search-word-data-structure-design/
// Author: github.com/lzl124631x
// Time:
//      WordDictionary: O(1)
//      addWord: O(1)
//      search: O(NW)
// Space: O(NW)
class WordDictionary {
    unordered_map<int, vector<string>> m;
public:
    void addWord(string word) {
        m[word.size()].push_back(word);
    }
    bool search(string word) {
        if (m.count(word.size()) == 0) return false;
        for (auto s : m[word.size()]) {
            int i = 0;
            for (; i < word.size(); ++i) {
                if (word[i] != '.' && word[i] != s[i]) break;
            }
            if (i == word.size()) return true;
        }
        return false;
    }
};