Implement Magic Dictionary Problem
Description
LeetCode Problem 676.
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 theMagicDictionaryclass:
- MagicDictionary()Initializes the object.
- void buildDict(String[]dictionary)Sets the data structurewith 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:
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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 indictionaryare distinct.
- 1 <=searchWord.length <= 100
- searchWordconsists of only lower-case English letters.
- buildDictwill be called only once before search.
- At most 100 calls will be made to search.
Sample C++ Code
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class MagicDictionary {
public:
vector<string> dict;
/** Initialize your data structure here. */
MagicDictionary() {
}
void buildDict(vector<string> dictionary) {
for (int i = 0; i < dictionary.size(); i ++) {
dict.push_back(dictionary[i]);
}
return;
}
bool checkDiff(string a, string b) {
if (a.size() != b.size())
return false;
int cnt_diff = 0;
for (int i = 0; i < a.size(); i ++) {
if (a[i] != b[i])
cnt_diff ++;
}
return (cnt_diff == 1);
}
bool search(string searchWord) {
bool exist = false;
for (int i = 0; i < dict.size(); i ++) {
if (checkDiff(searchWord, dict[i])) {
exist = true;
break;
}
}
return exist;
}
};
/**
* Your MagicDictionary object will be instantiated and called as such:
* MagicDictionary* obj = new MagicDictionary();
* obj->buildDict(dictionary);
* bool param_2 = obj->search(searchWord);
*/