676. Implement Magic Dictionary¶

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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.

Solution¶

import java.util.ArrayList;

class MagicDictionary {
    private ArrayList<String> res;
    public MagicDictionary() {
        res = new ArrayList<>();
    }

    public void buildDict(String[] dictionary) {
        for (String word : dictionary)
            res.add(word);
    }

    public boolean search(String searchWord) {
        for (String word : res) {
            if (ok(word, searchWord))
                return true;
        }
        return false;
    }

    private boolean ok(String word, String searchWord) {
        int n = word.length();
        int m = searchWord.length();
        if (n != m)
            return false;
        int count = 0;
        for (int i = 0; i < n; i++) {
            if (word.charAt(i) != searchWord.charAt(i))
                count++;
        }
        return count == 1;

    }
}

/**
    Your MagicDictionary object will be instantiated and called as such:
    MagicDictionary obj = new MagicDictionary();
    obj.buildDict(dictionary);
    boolean param_2 = obj.search(searchWord);
*/

Complexity Analysis¶

• Time Complexity: O(?)
• Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here