Add and Search Word — Data structure design

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 <= 25
• word in addWord consists of lowercase English letters.
• word in search consist of '.' or lowercase English letters.
• There will be at most 2 dots in word for search queries.
• At most 104 calls will be made to addWord and search.

Approach: Trie with DFS for Wildcards

Algorithm

1. Store words in a trie; each node has 26 children and an isEnd flag
2. addWord: insert each character along the trie, marking the last node as end
3. search: traverse recursively; when the current character is '.', try every child branch; otherwise follow the specific child. Return true if a full word exists for some branch

Time & Space Complexity

• Time Complexity: O(L) for addWord; O(26^d * L) worst case for search, where d is the number of dots
• Space Complexity: O(N * L) - the trie nodes for N words of average length L

Java Implementation

public class WordDictionary {

    private static class TrieNode {
        TrieNode[] children = new TrieNode[26];
        boolean isEnd;
    }

    private final TrieNode root;

    public WordDictionary() {
        root = new TrieNode();
    }

    public void addWord(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            int index = c - 'a';
            if (node.children[index] == null) {
                node.children[index] = new TrieNode();
            }
            node = node.children[index];
        }
        node.isEnd = true;
    }

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

    private boolean search(TrieNode node, String word, int pos) {
        if (node == null) {
            return false;
        }
        if (pos == word.length()) {
            return node.isEnd;
        }

        char c = word.charAt(pos);
        if (c == '.') {
            for (TrieNode child : node.children) {
                if (search(child, word, pos + 1)) {
                    return true;
                }
            }
            return false;
        }

        return search(node.children[c - 'a'], word, pos + 1);
    }

    // Test method
    public static void main(String[] args) {
        WordDictionary wordDictionary = new WordDictionary();
        wordDictionary.addWord("bad");
        wordDictionary.addWord("dad");
        wordDictionary.addWord("mad");
        System.out.println(wordDictionary.search("pad")); // return false
        System.out.println(wordDictionary.search("bad")); // return true
        System.out.println(wordDictionary.search(".ad")); // return true
        System.out.println(wordDictionary.search("b..")); // return true
    }
}

Example Walkthrough

After adding “bad”, “dad”, “mad”:

1. search("pad"): the p child of the root does not exist -> false
2. search("bad"): follows b -> a -> d, final node isEnd = true -> true
3. search(".ad"): try all 26 children of root; only b, d, m branches exist, and b.ad, d.ad, m.ad all match -> true
4. search("b.."): at the second position try all children of node b, eventually “bad” matches -> true

Key Points

1. Trie Storage: Prefix sharing keeps the dictionary compact and enables character-by-character matching
2. Wildcard Branching: A '.' forces recursion over all 26 children
3. Recursive Search: The search index tracks progress instead of slicing strings
4. End Marker: isEnd distinguishes complete words from prefixes
5. Bounded Dots: The constraint of at most 2 dots keeps the exponential factor small