# 676. Implement Magic Dictionary LeetCode Solution | Easy Approach

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Implement Magic Dictionary 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 Magic DictionaryDesign 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`.

Time: O(n)
Space: O(n)

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