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.

Implement Magic Dictionary Solutions

Time: O(n)
Space: O(n)

C++

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Java

 
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Python

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