745. Prefix and Suffix Search
Description
Design a special dictionary that searches the words in it by a prefix and a suffix.
Implement the WordFilter class:
WordFilter(string[] words)Initializes the object with thewordsin the dictionary.f(string pref, string suff)Returns the index of the word in the dictionary, which has the prefixprefand the suffixsuff. If there is more than one valid index, return the largest of them. If there is no such word in the dictionary, return-1.
Example 1:
Input
["WordFilter", "f"]
[[["apple"]], ["a", "e"]]
Output
[null, 0]
Explanation
WordFilter wordFilter = new WordFilter(["apple"]);
wordFilter.f("a", "e"); // return 0, because the word at index 0 has prefix = "a" and suffix = "e".
Constraints:
1 <= words.length <= 1041 <= words[i].length <= 71 <= pref.length, suff.length <= 7words[i],prefandsuffconsist of lowercase English letters only.- At most
104calls will be made to the functionf.
Solutions
Solution: Trie
- Time complexity: O(words.length*word[index].length2)
- Space complexity: O(words.length*word[index].length2)
JavaScript
js
/**
* @param {string[]} words
*/
const WordFilter = function (words) {
this.trie = new Map();
for (const [index, word] of words.entries()) {
let prefix = '';
for (let left = 0; left < word.length; left++) {
let current = this.trie;
prefix += word[left];
if (!current.has(prefix)) {
current.set(prefix, new Map());
}
current = current.get(prefix);
for (let right = 0; right < word.length; right++) {
const suffix = word.slice(right);
current.set(suffix, index);
}
}
}
};
/**
* @param {string} pref
* @param {string} suff
* @return {number}
*/
WordFilter.prototype.f = function (pref, suff) {
if (!this.trie.has(pref)) return -1;
const current = this.trie.get(pref);
return current.get(suff) ?? -1;
};
/**
* Your WordFilter object will be instantiated and called as such:
* var obj = new WordFilter(words)
* var param_1 = obj.f(pref,suff)
*/