Design Add and Search Words Data Structure

Problem

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 1:

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.

Solution

Data Structure:

• Trie (Prefix Tree): A tree-like data structure where each node represents a single character of a word. The root node is an entry point for all words. Each node has up to 26 children (one for each letter 'a' to 'z').

Adding a Word:

• Start at the root node.
• For each character in the word:
• Calculate the index of the character (character - 'a').
• If the corresponding child node is NULL, create a new TrieNode. Move to the child node.
• After processing all characters, mark the current node as a leaf node to indicate the end of a word.

Searching for a Word:

• Start at the root node.
• Use a recursive helper function searchHelper to handle the search:
• If you reach the end of the word, check if the current node is a leaf node.
• If the current character is '.', check all possible children nodes.
• If the current character is a regular letter, move to the corresponding child node.
• Return true if a matching word is found, otherwise false.

Code in Different Languages

C++

class TrieNode{
    public:
    TrieNode *children[26];
    bool isLeaf = false;

    TrieNode(){
        for(int i=0;i<26;i++) children[i] = NULL;
        isLeaf = false;
    }
};

TrieNode *root;

class WordDictionary {
public:
    WordDictionary() {
       root = new TrieNode() ;
    }
    
    void addWord(string word) {
        TrieNode *curr = root;
        for(int i=0;i<word.size();i++){
            if (word[i] < 'a' || word[i] > 'z') return;
            int index = word[i] - 'a';
            if(!curr->children[index]) curr->children[index] = new TrieNode();
            curr = curr->children[index];
        }
        curr->isLeaf = true;
    }
    
    // bool search(string word) {
    //     TrieNode *curr = root;
    //     for(int i=0;i<word.size();i++){
    //         if (word[i] < 'a' || word[i] > 'z') return false;
    //         int index = word[i] - 'a';
    //         if(!curr->children[index]) return false;
    //         curr = curr->children[index];
    //     }
    //     return curr->isLeaf;
    // }

    bool searchHelper(TrieNode* curr, string word, int index) {
    if (index == word.size()) return curr->isLeaf;
    if (word[index] == '.') {
        for (int i = 0; i < 26; i++) {
            if (curr->children[i] && searchHelper(curr->children[i], word, index + 1)) return true;
        }
        return false;
    } else {
        int idx = word[index] - 'a';
        if (!curr->children[idx]) return false;
        return searchHelper(curr->children[idx], word, index + 1);
    }
}

    bool search(string word) {
    return searchHelper(root, word, 0);
}

};

Java

import java.util.*;

class WordDictionary {
    Map<Integer, List<String>> map;

    public WordDictionary() {
        map = new HashMap<>();
    }

    public void addWord(String word) {
        int index = word.length();
        if (!map.containsKey(index)) {
            List<String> list = new ArrayList<>();
            list.add(word);
            map.put(index, list);
        } else {
            map.get(index).add(word);
        }
    }

    public boolean search(String word) {
        int index = word.length();
        if (!map.containsKey(index)) {
            return false;
        }

        List<String> list = map.get(index);
        for (String s : list) {
            if (isSame(s, word)) {
                return true;
            }
        }
        return false;
    }

    private boolean isSame(String s, String word) {
        for (int i = 0; i < s.length(); i++) {
            if (word.charAt(i) != '.' && s.charAt(i) != word.charAt(i)) {
                return false;
            }
        }
        return true;
    }
}

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

Complexity Analysis

Time Complexity:

• Adding a Word: O(m), where m is the length of the word. Each character is processed once.
• Searching for a Word:
• Worst Case: O(m * 26^m), where m is the length of the word and all characters are '.' (wildcards). This is due to the recursive search through all possible children nodes at each level.

Space Complexity:

• TrieNode: O(1) per node.
• WordDictionary: O(n * m), where n is the number of words and m is the average length of the words.

References

• LeetCode Problem: Design Add and Search Words Data Structure

• Leetcode Solutions: Design Add and Search Words Data Structure