676. Implement Magic Dictionary

Problem

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 <=&nbsp;dictionary.length <= 100

• 1 <=&nbsp;dictionary[i].length <= 100

• dictionary[i] consists of only lower-case English letters.

• All the strings in dictionary are distinct.

• 1 <=&nbsp;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 (Java)

class MagicDictionary {
    private String[] dictionaryWords;

    public MagicDictionary() {
        dictionaryWords = null;
    }

    public void buildDict(String[] dictionary) {
        dictionaryWords = dictionary;
    }

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

    private boolean isOffByOneLetter(String word, String searchWord) {
        if (isDifferentLengths(word, searchWord) || word.equals(searchWord)) {
            return false;
        }
        int numDifferentLetters = 0;
        for (int i = 0; i < word.length(); i++) {
            if (isNotTheSameLetter(word.charAt(i), searchWord.charAt(i))) {
                numDifferentLetters++;
            }
            if (numDifferentLetters > 1) {
                return false;
            }
        }
        return numDifferentLetters == 1;
    }

    private boolean isDifferentLengths(String word, String searchWord) {
        return word.length() != searchWord.length();
    }

    private boolean isNotTheSameLetter(char c1, char c2) {
        return c1 != c2;
    }
}

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

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).