# Leetcode Design Add and Search Words Data Structure problem solution

In this Leetcode Design Add and Search Words Data Structure problem solution we need to Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

1. WordDictionary() Initializes the object.
2. void addWord(word) Adds word to the data structure, it can be matched later.
3. 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.

## Problem solution in Python.

```class WordDictionary(object):

def __init__(self):

self.parser = collections.defaultdict(set)

def search(self, word):

if len(word) not in self.parser:
return False
for x in self.parser[len(word)]:
if all(True if l == "." else l == x[i] for i,l in enumerate(word)):
return True
return False
```

## Problem solution in Java.

```class WordDictionary {
class TrieNode {
TrieNode[] children;
boolean isWord;

private final int N = 26;

TrieNode() {
children = new TrieNode[N];
}
}
TrieNode root;

public WordDictionary() {
root = new TrieNode();
root.isWord = true;
}

TrieNode cur = root;
for (char c : word.toCharArray()) {
if (cur.children[c - 'a'] == null) {
cur.children[c - 'a'] = new TrieNode();
}
cur = cur.children[c - 'a'];
}
cur.isWord = true;
}

public boolean search(String word) {
return dfsSearch(root, word, 0);
}

private boolean dfsSearch(TrieNode cur, String word, int i) {
if (cur == null) return false;
if (i == word.length()) return cur.isWord;
char c = word.charAt(i);
if (c == '.') {
for (int j = 0; j < 26; j++) {
if (dfsSearch(cur.children[j], word, i + 1)) {
return true;
}
}
return false;
} else {
return dfsSearch(cur.children[c - 'a'], word, i + 1);
}
}
}
```

## Problem solution in C++.

```class WordDictionary {
private:
vector<vector<string>>v;
public:

WordDictionary() : v(100002) {}

v[word.size()].push_back(word);
}

bool search(string word) {
int n = word.size();
for(int i=0;i<v[n].size();i++){
int c=0;
for(int j=0;j<n;j++){
if(v[n][i][j]==word[j] || word[j]=='.')
c++;
else
break;
}
if(c==n)
{
return 1;
}
}
return 0;
}
};
```

## Problem solution in C.

```typedef struct trie_node {
struct trie_node *sub_chars[26];
bool EOW;
} WordDictionary;

WordDictionary* wordDictionaryCreate() {
WordDictionary *obj = (WordDictionary*)calloc(1,sizeof(WordDictionary));
return obj;
}

void wordDictionaryAddWord(WordDictionary* obj, char * word) {
while((*word)) {
if(obj->sub_chars[(*word)-'a']==NULL) {
obj->sub_chars[(*word)-'a'] = (WordDictionary*)calloc(1,sizeof(WordDictionary));
}
obj = obj->sub_chars[(*word)-'a'];
word++;
}
obj->EOW = true;
}

bool wordDictionarySearch(WordDictionary* obj, char * word) {
while(*word) {
if( (*word) =='.') {
for(int i =0;i<26;i++) {
if(!obj->sub_chars[i]) continue;
bool a = wordDictionarySearch(obj->sub_chars[i],word+1);
if(a) return true;
}
return false;
}
else if(!obj->sub_chars[(*word)-'a']) return false;
obj = obj->sub_chars[(*word)-'a'];
word++;
}
if(obj->EOW)return true;
else return false;
}

void wordDictionaryFree(WordDictionary* obj) {
for(int i=0;i<26;i++) {
if(!obj->sub_chars[i]) continue;
wordDictionaryFree(obj->sub_chars[i]);
}
free(obj);
}
```