642 Design Search Autocomplete System

1. Question

Design a search autocomplete system for a search engine. Users may input a sentence (at least one word and end with a special character'#'). For each character they type except '#', you need to return the top 3 historical hot sentences that have prefix the same as the part of sentence already typed. Here are the specific rules:
  1. 1.
    The hot degree for a sentence is defined as the number of times a user typed the exactly same sentence before.
  2. 2.
    The returned top 3 hot sentences should be sorted by hot degree (The first is the hottest one). If several sentences have the same degree of hot, you need to use ASCII-code order (smaller one appears first).
  3. 3.
    If less than 3 hot sentences exist, then just return as many as you can.
  4. 4.
    When the input is a special character, it means the sentence ends, and in this case, you need to return an empty list.
Your job is to implement the following functions:
The constructor function:
AutocompleteSystem(String[] sentences, int[] times):This is the constructor. The input is historical data.Sentencesis a string array consists of previously typed sentences.Timesis the corresponding times a sentence has been typed. Your system should record these historical data.
Now, the user wants to input a new sentence. The following function will provide the next character the user types:
List<String> input(char c):The inputcis the next character typed by the user. The character will only be lower-case letters ('a'to'z'), blank space (' ') or a special character ('#'). Also, the previously typed sentence should be recorded in your system. The output will be the top 3 historical hot sentences that have prefix the same as the part of sentence already typed.
Example: Operation: AutocompleteSystem(["i love you", "island","ironman", "i love leetcode"], [5,3,2,2]) The system have already tracked down the following sentences and their corresponding times: "i love you":5times "island":3times "ironman":2times "i love leetcode":2times Now, the user begins another search:
Operation: input('i') Output: ["i love you", "island","i love leetcode"] Explanation: 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.
Operation: input(' ') Output: ["i love you","i love leetcode"] Explanation: There are only two sentences that have prefix"i ".
Operation: input('a') Output: [] Explanation: There are no sentences that have prefix"i a".
Operation:input('#') Output:[] Explanation: 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.
Note:
  1. 1.
    The input sentence will always start with a letter and end with '#', and only one blank space will exist between two words.
  2. 2.
    The number of complete sentences that to be searched won't exceed 100. The length of each sentence including those in the historical data won't exceed 100.
  3. 3.
    Please use double-quote instead of single-quote when you write test cases even for a character input.
  4. 4.
    Please remember to RESET your class variables declared in class AutocompleteSystem, as static/class variables are
    persisted across multiple test cases. Please see here for more details.

2. Implementation

(1) Trie + PriorityQueue
1
class AutocompleteSystem {
2
class TrieNode {
3
char val;
4
boolean isWord;
5
Map<String, Integer> count;
6
TrieNode[] childNode;
7
8
public TrieNode(char val) {
9
this.val = val;
10
count = new HashMap<>();
11
childNode = new TrieNode[27];
12
}
13
}
14
15
class Trie {
16
TrieNode root;
17
18
public Trie() {
19
root = new TrieNode(' ');
20
}
21
22
public void insert(String sentence, int frequency) {
23
TrieNode curNode = root;
24
25
for (char c : sentence.toCharArray()) {
26
int index = c == ' ' ? 26 : c - 'a';
27
28
if (curNode.childNode[index] == null) {
29
curNode.childNode[index] = new TrieNode(c);
30
}
31
curNode = curNode.childNode[index];
32
curNode.count.put(sentence, curNode.count.getOrDefault(sentence, 0) + frequency);
33
}
34
}
35
36
public TrieNode search(String prefix) {
37
TrieNode curNode = root;
38
39
for (char c : prefix.toCharArray()) {
40
int index = c == ' ' ? 26 : c - 'a';
41
42
if (curNode.childNode[index] == null) {
43
return null;
44
}
45
curNode = curNode.childNode[index];
46
}
47
return curNode;
48
}
49
}
50
51
class WordInfo implements Comparable<WordInfo> {
52
String word;
53
int count;
54
55
public WordInfo(String word, int count) {
56
this.word = word;
57
this.count = count;
58
}
59
60
public int compareTo(WordInfo that) {
61
return this.count == that.count ? this.word.compareTo(that.word) : that.count - this.count;
62
}
63
}
64
65
Trie trie;
66
String prefix;
67
68
public AutocompleteSystem(String[] sentences, int[] times) {
69
trie = new Trie();
70
prefix = "";
71
72
for (int i = 0; i < sentences.length; i++) {
73
trie.insert(sentences[i], times[i]);
74
}
75
}
76
77
public List<String> input(char c) {
78
List<String> res = new ArrayList<>();
79
80
if (c == '#') {
81
trie.insert(prefix, 1);
82
prefix = "";
83
return res;
84
}
85
86
prefix = prefix + c;
87
TrieNode node = trie.search(prefix);
88
89
if (node == null) {
90
return res;
91
}
92
93
PriorityQueue<WordInfo> maxHeap = new PriorityQueue<>();
94
95
for (String key : node.count.keySet()) {
96
maxHeap.add(new WordInfo(key, node.count.get(key)));
97
}
98
99
for (int i = 0; i < 3 && !maxHeap.isEmpty(); i++) {
100
res.add(maxHeap.remove().word);
101
}
102
return res;
103
}
104
}
105
106
/**
107
* Your AutocompleteSystem object will be instantiated and called as such:
108
* AutocompleteSystem obj = new AutocompleteSystem(sentences, times);
109
* List<String> param_1 = obj.input(c);
110
*/
Copied!

3. Time & Space Complexity

Trie + PriorityQueue:
时间复杂度
(1) Autocomplete: O(nL),n为sentence的个数, L为sentence的平均长度.
(2) input: O(q + klogk), q为目前行成prefix的长度, k为node里存有的sentence的个数
空间复杂度O(nL + k)