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Trie - Question 2

 Leetcode 211. Design Add and Search Words Data Structure

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.

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.


Analysis:

This question asks to build a word dictionary, which essentially a trie.

So we can define a node structure which has 26 children and a Boolean to label whether a string (or prefix) a word or not.

With this node structure, we can build a trie containing all the words. For each letter in the word, we move down one layer in the trie.

After building the trie, we can do the search, to determine whether a string is in the trie or not. Since the input string has wildcard (*), so recursive solution becomes suitable. The key to recursive solution is: the same question but a smaller size.


See the code below:


class WordDictionary {
struct node {
    vector<node*> nds;
    bool isW;
    node() {
        nds.resize(26);
        isW = false;
    }
};
public:
    /** Initialize your data structure here. */
    WordDictionary() {
        root = new node();
    }
    
    void addWord(string word) {
        node* p = root;
        for(auto &a : word) {
            int id = a - 'a';
            if(!p->nds[id]) p->nds[id] = new node();
            p = p->nds[id];
        }
        p->isW = true;
    }
    
    bool search(string word) {
        return isV(root, word);
    }
private:
    node* root;
    bool isV(node* p, string word) {
        if(word.empty()) return p->isW;
        if(word[0] != '.') {
            int id = word[0] - 'a';
            if(!p->nds[id]) return false;
            return isV(p->nds[id], word.substr(1));
        }
        for(auto &a : p->nds) {
            if(a && isV(a, word.substr(1))) return true;
        }
        return false;
    }
};

/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary* obj = new WordDictionary();
 * obj->addWord(word);
 * bool param_2 = obj->search(word);
 */


Upper Layer

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