# Implement Magic Dictionary

Implement a magic directory with `buildDict`, and `search` methods.

For the method `buildDict`, you'll be given a list of non-repetitive words to build a dictionary.

For the method `search`, you'll be given a word, and judge whether if you modify **exactly** one character into **another** character in this word, the modified word is in the dictionary you just built.

**Example 1:**\


```
Input: buildDict(["hello", "leetcode"]), Output: Null
Input: search("hello"), Output: False
Input: search("hhllo"), Output: True
Input: search("hell"), Output: False
Input: search("leetcoded"), Output: False
```

**Note:**\


1. You may assume that all the inputs are consist of lowercase letters `a-z`.
2. For contest purpose, the test data is rather small by now. You could think about highly efficient algorithm after the contest.
3. Please remember to **RESET** your class variables declared in class MagicDictionary, as static/class variables are **persisted across multiple test cases**. Please see [here](https://leetcode.com/faq/#different-output) for more details.

```java
class MagicDictionary {

    Map<String, List<int[]>> map;

    public MagicDictionary() {
        map = new HashMap<>();
    }

    /** Build a dictionary through a list of words */
    public void buildDict(String[] dict) {
        for (String s : dict) {
            for (int i = 0; i < s.length(); i++) {
                String key = s.substring(0, i) + s.substring(i + 1);
                int[] pair = new int[] { i, s.charAt(i) };

                List<int[]> val = map.getOrDefault(key, new ArrayList<int[]>());
                val.add(pair);

                map.put(key, val);
            }
        }
    }

    /**
     * Returns if there is any word in the trie that equals to the given word after
     * modifying exactly one character
     */
    public boolean search(String word) {
        for (int i = 0; i < word.length(); i++) {
            String key = word.substring(0, i) + word.substring(i + 1);
            if (map.containsKey(key)) {
                for (int[] pair : map.get(key))
                    if (pair[0] == i && pair[1] != word.charAt(i))
                        return true;
            }
        }
        return false;
    }
}
```
