# [LeetCode] 676. Implement Magic Dictionary 实现神奇字典

2021年09月15日 阅读数：1

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

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

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.python

Example 1:函数

```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:post

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 for more details.

Java:this

```class MagicDictionary {

Map<String, List<int[]>> map = new HashMap<>();
/** Initialize your data structure here. */
public MagicDictionary() {
}

/** 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[]>());

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;
}
}
```

Python:url

```class MagicDictionary(object):
def _candidates(self, word):
for i in xrange(len(word)):
yield word[:i] + '*' + word[i+1:]

def buildDict(self, words):
self.words = set(words)
self.near = collections.Counter(cand for word in words
for cand in self._candidates(word))

def search(self, word):
return any(self.near[cand] > 1 or
self.near[cand] == 1 and word not in self.words
for cand in self._candidates(word))
```

Python:code

```# Time:  O(n), n is the length of the word
# Space: O(d)

import collections

class MagicDictionary(object):

def __init__(self):
"""
"""
_trie = lambda: collections.defaultdict(_trie)
self.trie = _trie()

def buildDict(self, dictionary):
"""
Build a dictionary through a list of words
:type dictionary: List[str]
:rtype: void
"""
for word in dictionary:
reduce(dict.__getitem__, word, self.trie).setdefault("_end")

def search(self, word):
"""
Returns if there is any word in the trie that equals to the given word after modifying exactly one character
:type word: str
:rtype: bool
"""
def find(word, curr, i, mistakeAllowed):
if i == len(word):
return "_end" in curr and not mistakeAllowed

if word[i] not in curr:
return any(find(word, curr[c], i+1, False) for c in curr if c != "_end") \
if mistakeAllowed else False

if mistakeAllowed:
return find(word, curr[word[i]], i+1, True) or \
any(find(word, curr[c], i+1, False) \
for c in curr if c not in ("_end", word[i]))
return find(word, curr[word[i]], i+1, False)

return find(word, self.trie, 0, True)　　```

C++:htm

```class MagicDictionary {
public:
/** Initialize your data structure here. */
MagicDictionary() {}

/** Build a dictionary through a list of words */
void buildDict(vector<string> dict) {
for (string word : dict) {
m[word.size()].push_back(word);
}
}

/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
bool search(string word) {
for (string str : m[word.size()]) {
int cnt = 0, i = 0;
for (; i < word.size(); ++i) {
if (word[i] == str[i]) continue;
if (word[i] != str[i] && cnt == 1) break;
++cnt;
}
if (i == word.size() && cnt == 1) return true;
}
return false;
}

private:
unordered_map<int, vector<string>> m;
};　　```

C++:

```class MagicDictionary {
public:
/** Initialize your data structure here. */
MagicDictionary() {}

/** Build a dictionary through a list of words */
void buildDict(vector<string> dict) {
for (string word : dict) s.insert(word);
}

/** Returns if there is any word in the trie that equals to the given word after modifying exactly one character */
bool search(string word) {
for (int i = 0; i < word.size(); ++i) {
char t = word[i];
for (char c = 'a'; c <= 'z'; ++c) {
if (c == t) continue;
word[i] = c;
if (s.count(word)) return true;
}
word[i] = t;
}
return false;
}

private:
unordered_set<string> s;
};
```

[LeetCode] 208. Implement Trie (Prefix Tree) 实现字典树(前缀树)

720. Longest Word in Dictionary