642. Design Search Autocomplete System
Design a search autocomplete system for a search engine. Users may input a sentence (at least one word and end with a special character '#'
).
You are given a string array sentences
and an integer array times
both of length n
where sentences[i]
is a previously typed sentence and times[i]
is the corresponding number of times the sentence was typed. For each input character except '#'
, return the top 3
historical hot sentences that have the same prefix as the part of the sentence already typed.
Here are the specific rules:
- The hot degree for a sentence is defined as the number of times a user typed the exactly same sentence before.
- The returned top
3
hot sentences should be sorted by hot degree (The first is the hottest one). If several sentences have the same hot degree, use ASCII-code order (smaller one appears first). - If less than
3
hot sentences exist, return as many as you can. - When the input is a special character, it means the sentence ends, and in this case, you need to return an empty list.
Implement the AutocompleteSystem
class:
-
AutocompleteSystem(String[] sentences, int[] times)
Initializes the object with thesentences
andtimes
arrays. -
List<String> input(char c)
This indicates that the user typed the characterc
. - Returns an empty array
[]
ifc == '#'
and stores the inputted sentence in the system. - Returns the top
3
historical hot sentences that have the same prefix as the part of the sentence already typed. If there are fewer than3
matches, return them all.
Example 1:
Input
["AutocompleteSystem", "input", "input", "input", "input"]
[[["i love you", "island", "iroman", "i love leetcode"], [5, 3, 2, 2]], ["i"], [" "], ["a"], ["#"]]
Output
[null, ["i love you", "island", "i love leetcode"], ["i love you", "i love leetcode"], [], []]
Explanation
AutocompleteSystem obj = new AutocompleteSystem(["i love you", "island", "iroman", "i love leetcode"], [5, 3, 2, 2]);
obj.input("i"); // return ["i love you", "island", "i love leetcode"]. There are four sentences that have prefix "i". Among them, "ironman" and "i love leetcode" have same hot degree. Since ' ' has ASCII code 32 and 'r' has ASCII code 114, "i love leetcode" should be in front of "ironman". Also we only need to output top 3 hot sentences, so "ironman" will be ignored.
obj.input(" "); // return ["i love you", "i love leetcode"]. There are only two sentences that have prefix "i ".
obj.input("a"); // return []. There are no sentences that have prefix "i a".
obj.input("#"); // return []. The user finished the input, the sentence "i a" should be saved as a historical sentence in system. And the following input will be counted as a new search.
Constraints:
n == sentences.length
n == times.length
1 <= n <= 100
1 <= sentences[i].length <= 100
1 <= times[i] <= 50
c
is a lowercase English letter, a hash'#'
, or space' '
.- Each tested sentence will be a sequence of characters
c
that end with the character'#'
. - Each tested sentence will have a length in the range
[1, 200]
. - The words in each input sentence are separated by single spaces.
- At most
5000
calls will be made toinput
.
Solution:
Do this first:
208. Implement Trie (Prefix Tree)
class AutocompleteSystem {
static class TrieNode {
Map<Character, TrieNode> children = new HashMap<>();
Map<String, Integer> frequencyMap = new HashMap<>();
boolean isEnd = false;
}
static class Trie {
TrieNode root;
Trie() {
root = new TrieNode();
}
public void insert(String sentence, int times) {
TrieNode cur = root;
for (char c : sentence.toCharArray()) {
cur.children.putIfAbsent(c, new TrieNode());
cur = cur.children.get(c);
cur.frequencyMap.put(sentence, cur.frequencyMap.getOrDefault(sentence, 0) + times);
}
cur.isEnd = true;
}
public Map<String, Integer> search(String prefix) {
TrieNode cur = root;
for (char c : prefix.toCharArray()) {
if (!cur.children.containsKey(c)) {
return new HashMap<>();
}
cur = cur.children.get(c);
}
return cur.frequencyMap;
}
}
private final Trie trie;
private final StringBuilder currentInput; // Stores the characters typed so far for the current input.
public AutocompleteSystem(String[] sentences, int[] times) {
trie = new Trie();
currentInput = new StringBuilder();
for (int i = 0; i < sentences.length; i++) {
trie.insert(sentences[i], times[i]);
}
}
public List<String> input(char c) {
List<String> result = new ArrayList<>();
if (c == '#') {
// Finalize the current input
String sentence = currentInput.toString();
trie.insert(sentence, 1);
currentInput.setLength(0);
// The setLength(0) method clears the StringBuilder, effectively resetting the currentInput to an empty string.
return result;
}
currentInput.append(c);
String prefix = currentInput.toString();
// Retrieve all matching sentences and sort them by hotness and lexicographical order
Map<String, Integer> frequencyMap = trie.search(prefix);
// PriorityQueue<String> pq = new PriorityQueue<>((a, b) -> (frequencyMap.get(b) - frequencyMap.get(a)));// maxheap
// 优先队列比较器:按频率降序,频率相同时按字典序升序
PriorityQueue<String> pq = new PriorityQueue<>((a, b) -> {
int freqCompare = frequencyMap.get(b).compareTo(frequencyMap.get(a));
return freqCompare != 0 ? freqCompare : a.compareTo(b);
});
for (String sentence : frequencyMap.keySet()) {
pq.offer(sentence);
}
for (int i = 0; i < 3 && !pq.isEmpty(); i++) {
result.add(pq.poll());
}
return result;
}
}
/**
* Your AutocompleteSystem object will be instantiated and called as such:
* AutocompleteSystem obj = new AutocompleteSystem(sentences, times);
* List<String> param_1 = obj.input(c);
*/
PriorityQueue<String> pq = new PriorityQueue<>(new Comparator<String>() {
@Override
public int compare(String a, String b) {
// Compare by frequency in descending order
int freqCompare = frequencyMap.get(b) - frequencyMap.get(a);
if (freqCompare != 0) {
return freqCompare; // If frequencies are different, use frequency comparison
}
// If frequencies are the same, compare lexicographically
return a.compareTo(b);
}
});
PriorityQueue<String> pq = new PriorityQueue<>((a, b) -> {
// Compare by frequency in descending order
int freqCompare = frequencyMap.get(b) - frequencyMap.get(a);
if (freqCompare != 0) {
return freqCompare; // If frequencies are different, use frequency comparison
}
// If frequencies are the same, compare lexicographically
return a.compareTo(b);
});