Problem Statement

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.

Problem Link

Design Add and Search Words Data Structure - LeetCode

Code

class WordDictionary {
    vector<WordDictionary*> children;
    bool isEndOfWord;
public:
    // Initialize your data structure here. 
    WordDictionary(): isEndOfWord(false){
        children = vector<WordDictionary*>(26, nullptr);
    }
    
    // Adds a word into the data structure. 
    void addWord(string word) {
        WordDictionary *curr = this;
        for(char c: word){
            if(curr->children[c - 'a'] == nullptr)
                curr->children[c - 'a'] = new WordDictionary();
            curr = curr->children[c - 'a'];
        }
        curr->isEndOfWord = true;
    }
    
    // Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. 
    bool search(string word) {
        WordDictionary *curr = this;
        for(int i = 0; i < word.length(); ++i){
            char c = word[i];
            if(c == '.'){
                for(auto ch: curr->children)
                    if(ch && ch->search(word.substr(i+1))) return true;
                return false;
            }
            if(curr->children[c - 'a'] == nullptr) return false;
            curr = curr->children[c - 'a'];
        }
        return curr && curr->isEndOfWord;
    }
};

• Time Complexity :- BigO(M) for well defined words, But in worse case BigO(M.26^N)
• Space Complexity :- BigO(1) for well defined words, But for worst case BigO(M)