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Java Solution for LeetCode algorithm problems, continually updating

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676\. Implement Magic Dictionary

Medium

Design a data structure that is initialized with a list of **different** words. Provided a string, you should determine if you can change exactly one character in this string to match any word in the data structure.

Implement the `MagicDictionary` class:

*   `MagicDictionary()` Initializes the object.
*   `void buildDict(String[] dictionary)` Sets the data structure with an array of distinct strings `dictionary`.
*   `bool search(String searchWord)` Returns `true` if you can change **exactly one character** in `searchWord` to match any string in the data structure, otherwise returns `false`.

**Example 1:**

**Input** 

["MagicDictionary", "buildDict", "search", "search", "search", "search"] [[], 

[["hello", "leetcode"]], ["hello"], ["hhllo"], ["hell"], ["leetcoded"]]

**Output:** [null, null, false, true, false, false]

**Explanation:** 

    MagicDictionary magicDictionary = new MagicDictionary(); 
    magicDictionary.buildDict(["hello", "leetcode"]); 
    magicDictionary.search("hello"); // return False 
    magicDictionary.search("hhllo"); // We can change the second 'h' to 'e' to match "hello" so we return True 
    magicDictionary.search("hell"); // return False 
    magicDictionary.search("leetcoded"); // return False

**Constraints:**

*   `1 <= dictionary.length <= 100`
*   `1 <= dictionary[i].length <= 100`
*   `dictionary[i]` consists of only lower-case English letters.
*   All the strings in `dictionary` are **distinct**.
*   `1 <= searchWord.length <= 100`
*   `searchWord` consists of only lower-case English letters.
*   `buildDict` will be called only once before `search`.
*   At most `100` calls will be made to `search`.