Design Add and Search Words Data Structure


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 <= 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 :



title-img 


                            Solution in C :

typedef struct Trie {
    struct Trie *children[26];
    bool isWord;
} Trie;

typedef struct {
    Trie *base;
} WordDictionary;

Trie* trieCreate() {
    Trie *base = (Trie*) malloc(sizeof(Trie));
    for (int i = 0; i < 26; i++) {
        base->children[i] = NULL;
    }    
    base->isWord = false;
    return base;
}

WordDictionary* wordDictionaryCreate() {
	WordDictionary *obj = malloc(sizeof(WordDictionary));
	obj->base = trieCreate();
	return obj;
}

void trieInsert(Trie* trie, char * word) {
    int n = strlen(word);
    Trie* curr = trie;
    for (int i = 0; i < n; i++) {
        int index = word[i]-'a';
        if (curr->children[index] == NULL) {
            curr->children[index] = trieCreate();
        } 
        curr = curr->children[index];
        if (i == n-1) {
            curr->isWord = true;
        }
    }
}

void wordDictionaryAddWord(WordDictionary* obj, char * word) {
	trieInsert(obj->base, word);
}

bool trieSearch(Trie* trie, char * word) {
    int i, j, index, n = strlen(word);
    Trie* curr = trie;
    for (i = 0; i < n; i++) {
		if (word[i] == '.') {
			for (j = 0; j < 26; ++j) {
				if (curr->children[j] != NULL) {
					if (trieSearch(curr->children[j], &word[i+1])) {
						return true;
					}
				}
			}
			return false;
		} else {
			index = word[i]-'a';
			if (curr->children[index] == NULL) {
				return false;
			}
			curr = curr->children[index];
		}
	}
	return curr->isWord;
}

bool wordDictionarySearch(WordDictionary* obj, char * word) {
	return trieSearch(obj->base, word);
}

void trieFree(Trie* obj) {
	if (obj != NULL) {
		for (int i=0; i < 26; ++i) {
			trieFree(obj->children[i]);
		}
		free(obj);
	}
}

void wordDictionaryFree(WordDictionary* obj) {
    trieFree(obj->base);
    free(obj);
}

/**
 * Your WordDictionary struct will be instantiated and called as such:
 * WordDictionary* obj = wordDictionaryCreate();
 * wordDictionaryAddWord(obj, word);
 
 * bool param_2 = wordDictionarySearch(obj, word);
 
 * wordDictionaryFree(obj);
*/
                        


                        Solution in C++ :

class TrieNode {
public:
    TrieNode* children[26];
    bool isWordCompleted;

    TrieNode() {
        memset(children, 0, sizeof(children));
        isWordCompleted = false;
    }
};

class WordDictionary {
public:
    TrieNode* root;
    WordDictionary() {
        root = new TrieNode();
    }
    
    void addWord(string word) {
        TrieNode* newRoot = root;
        for (char ch : word) {
            int alphabetIndex = ch - 'a';
            if (newRoot -> children[alphabetIndex] == NULL) {
                newRoot -> children[alphabetIndex] = new TrieNode();
            }
            newRoot = newRoot -> children[alphabetIndex];
        }
        newRoot -> isWordCompleted = true;
    }
    
    bool searchHelper(string word, int index, TrieNode* newRoot) {
        if (index == word.length())
            return newRoot -> isWordCompleted;
        char ch = word[index];
        if (ch == '.') {
            for (int i = 0; i < 26; i++) {
                if (newRoot -> children[i] != NULL && searchHelper(word, index + 1, newRoot -> children[i])) {
                    return true;
                }
            }
            return false;
        }
        else {
            if (newRoot -> children[ch - 'a'] == NULL) {
                return false;
            }
            return (searchHelper(word, index + 1, newRoot -> children[ch - 'a']));
        }
    }

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

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


                        Solution in Java :

class TrieNode {
    TrieNode[] children;
    boolean isWordCompleted;

    TrieNode() {
        children = new TrieNode[26];
        isWordCompleted = false;
    }
}

class WordDictionary {
    TrieNode root;

    WordDictionary() {
        root = new TrieNode();
    }
    
    void addWord(String word) {
        TrieNode newRoot = root;
        for (char ch : word.toCharArray()) {
            int alphabetIndex = ch - 'a';
            if (newRoot.children[alphabetIndex] == null) {
                newRoot.children[alphabetIndex] = new TrieNode();
            }
            newRoot = newRoot.children[alphabetIndex];
        }
        newRoot.isWordCompleted = true;
    }
    
    boolean searchHelper(String word, int index, TrieNode newRoot) {
        if (index == word.length())
            return newRoot.isWordCompleted;
        char ch = word.charAt(index);
        if (ch == '.') {
            for (int i = 0; i < 26; i++) {
                if (newRoot.children[i] != null && searchHelper(word, index + 1, newRoot.children[i])) {
                    return true;
                }
            }
            return false;
        }
        else {
            if (newRoot.children[ch - 'a'] == null) {
                return false;
            }
            return searchHelper(word, index + 1, newRoot.children[ch - 'a']);
        }
    }

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


                        Solution in Python : 
                            
class TrieNode:
    def __init__(self):
        self.children = [None] * 26
        self.isWordCompleted = False

class WordDictionary:
    def __init__(self):
        self.root = TrieNode()
    
    def addWord(self, word: str) -> None:
        newRoot = self.root
        for ch in word:
            alphabetIndex = ord(ch) - ord('a')
            if newRoot.children[alphabetIndex] is None:
                newRoot.children[alphabetIndex] = TrieNode()
            newRoot = newRoot.children[alphabetIndex]
        newRoot.isWordCompleted = True
    
    def searchHelper(self, word: str, index: int, newRoot: TrieNode) -> bool:
        if index == len(word):
            return newRoot.isWordCompleted
        ch = word[index]
        if ch == '.':
            for i in range(26):
                if newRoot.children[i] is not None and self.searchHelper(word, index + 1, newRoot.children[i]):
                    return True
            return False
        else:
            alphabetIndex = ord(ch) - ord('a')
            if newRoot.children[alphabetIndex] is None:
                return False
            return self.searchHelper(word, index + 1, newRoot.children[alphabetIndex])

    def search(self, word: str) -> bool:
        return self.searchHelper(word, 0, self.root)


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