Given many words, words[i] has weight i.
Design a class WordFilter that supports one function, WordFilter.f(String prefix, String suffix). It will return the word with given prefix and suffix with maximum weight. If no word exists, return -1.
Examples:
Input:
WordFilter(["apple"])
WordFilter.f("a", "e") // returns 0
WordFilter.f("b", "") // returns -1
Note:
words has length in range [1, 15000].
For each test case, up to words.length queries WordFilter.f may be made.
words[i] has length in range [1, 10].
prefix, suffix have lengths in range [0, 10].
words[i] and prefix, suffix queries consist of lowercase letters only.
class WordFilter {
TrieNode trie;
public WordFilter(String[] words) {
trie = new TrieNode();
for (int weight = 0; weight < words.length; ++weight) {
String word = words[weight] + "{";
for (int i = 0; i < word.length(); ++i) {
TrieNode cur = trie;
cur.weight = weight;
for (int j = i; j < 2 * word.length() - 1; ++j) {
int k = word.charAt(j % word.length()) - 'a';
if (cur.children[k] == null)
cur.children[k] = new TrieNode();
cur = cur.children[k];
cur.weight = weight;
}
}
}
}
public int f(String prefix, String suffix) {
TrieNode cur = trie;
for (char letter: (suffix + '{' + prefix).toCharArray()) {
if (cur.children[letter - 'a'] == null) return -1;
cur = cur.children[letter - 'a'];
}
return cur.weight;
}
}
class TrieNode {
TrieNode[] children;
int weight;
public TrieNode() {
children = new TrieNode[27];
weight = 0;
}
}
/**
* Your WordFilter object will be instantiated and called as such:
* WordFilter obj = new WordFilter(words);
* int param_1 = obj.f(prefix,suffix);
*/
