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.

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.

Solution:

public class WordDictionary {

    public class TrieNode {
        public TrieNode[] children = new TrieNode[26];
        public boolean isWord;
    }
    
    private TrieNode root = new TrieNode();

    // Adds a word into the data structure.
    public void addWord(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            if (node.children[c - 'a'] == null) {
                node.children[c - 'a'] = new TrieNode();
            }
            node = node.children[c - 'a'];
        }
        node.isWord = true;
    }

 
    public boolean search(String word) {
        return match(word.toCharArray(), 0, root);
    }
    
    private boolean match(char[] chs, int k, TrieNode node) {
        if (k == chs.length) {
            return node.isWord;
        }
        if (chs[k] == '.') {
            for (int i = 0; i < node.children.length; i++) {
                if (node.children[i] != null && match(chs, k + 1, node.children[i])) {
                    return true;
                }
            }
        } else {
            return node.children[chs[k] - 'a'] != null && match(chs, k + 1, node.children[chs[k] - 'a']);
        }
        return false;
    }
}