# 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.

For example:

``````addWord("bad")
search("b..") -> true
``````

Note:

You may assume that all words are consist of lowercase letters a-z.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

## Solution

``````public class WordDictionary {
static class TrieNode {
// Initialize your data structure here.
TrieNode[] children = new TrieNode[26];
int count = 0;
public TrieNode() {
}

TrieNode safe(int i){
if(children[i] == null){
children[i] = new TrieNode();
}
return children[i];
}

int index(char c){
return (int)(c - 'a');
}

void insert(char[] word, int st, int len){
if(len == 0){
this.count++;
return;
}
TrieNode t = safe(index(word[st]));
t.insert(word, st + 1, len - 1);
}

boolean search(char[] word, int st, int len){
if(len == 0){
return this.count > 0;
}
if(word[st] == '.'){
for(TrieNode t : children){
if(t != null){
if(t.search(word, st + 1, len - 1)){
return true;
}
}
}
return false;
}
TrieNode t = children[index(word[st])];
if(t == null){
return false;
}
return t.search(word, st + 1, len - 1);
}
}
TrieNode root = new TrieNode();

// Adds a word into the data structure.
root.insert(word.toCharArray(), 0, word.length());
}

// 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 root.search(word.toCharArray(), 0, word.length());
}
}

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();