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Formatted question description: https://leetcode.ca/all/676.html

676. Implement Magic Dictionary (Medium)

Implement a magic directory with buildDict, and search methods.

For the method buildDict, you'll be given a list of non-repetitive words to build a dictionary.

For the method search, you'll be given a word, and judge whether if you modify exactly one character into another character in this word, the modified word is in the dictionary you just built.

Example 1:

Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False


  1. You may assume that all the inputs are consist of lowercase letters a-z.
  2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
  3. Please remember to RESET your class variables declared in class MagicDictionary, as static/class variables are persisted across multiple test cases. Please see here for more details.

Google, Amazon

Related Topics:
Hash Table, Trie

Solution 1. Trie

// OJ: https://leetcode.com/problems/implement-magic-dictionary/
// Time:
//   buildDict: O(W) where W is the size of all content in word.
//   search: O(W) since we only branch once in backtracking, it's not O(26^W)
// Space: O(W)
class TrieNode {
    TrieNode *next[26] = {};
    bool isWord = false;
class Trie {
    TrieNode root;
    void insert(string &word) {
        auto node = &root;
        for (char c : word) {
            if (!node->next[c - 'a']) node->next[c - 'a'] = new TrieNode();
            node = node->next[c - 'a'];
        node->isWord = true;
class MagicDictionary {
    Trie trie;
    bool search(string &word, int index, int diff, TrieNode *node) {
        if (diff > 1) return false;
        if (index == word.size()) return node->isWord && diff == 1;
        for (int i = 0; i < 26; ++i) {
            if (!node->next[i]) continue;
            if (search(word, index + 1, diff + (word[index] - 'a' == i ? 0 : 1), node->next[i])) return true;
        return false;
    void buildDict(vector<string> dict) {
        for (auto &word : dict) trie.insert(word);
    bool search(string word) {
        return search(word, 0, 0, &trie.root);

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