problemhardalgorithmsleetcode-642leetcode 642leetcode642

Design Search Autocomplete System

HardUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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. The hot degree for a sentence is defined as the number of times a user typed the exactly same sentence before.
  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. If less than 3 hot sentences exist, then just return as many as you can.
  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. Sentences is a string array consists of previously typed sentences. Times is 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 input c is 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.

Examples

**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"` : `5` times
`"island"` : `3` times
`"ironman"` : `2` times
`"i love leetcode"` : `2` times

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. The input sentence will always start with a letter and end with '#', and only one blank space will exist between two words.
  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. Please use double-quote instead of single-quote when you write test cases even for a character input.
  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.

Solution

Method 1 - Using Trie

Here is the approach:

  1. TrieNode Class:

    • Each node in the Trie contains:
      • A freq dictionary mapping a sentence to its frequency.
      • A children dictionary mapping characters to their respective Trie nodes.
      • An is_end flag to mark the end of a sentence.

  2. AutocompleteSystem Class:

    • Initialise the Trie with the sentences and their corresponding times by calling _insert on each sentence.
    • Maintain a sentence buffer (sb) and current Trie pointer (curr) during input processing.

  3. Input Method:

    • Append characters to the buffer while traversing or creating Trie nodes.
    • When encountering '#', store the completed sentence in the Trie with a frequency of 1 and reset the input state.
    • Fetch top 3 sentences matching the prefix using _find_top_k.

  4. Top-K Search:

    • Use a min heap (heapq) to keep track of the top 3 sentences based on both frequency and lexicographical order.
    • Sort and deliver results in descending order after popping from the heap.

  5. Reset:

    • Reset the sentence buffer and Trie pointer after a completed sentence is saved.

  6. Insert:

    • Traverse the Trie while creating missing nodes and updating frequency maps for nodes along the sentence's path.

Code

Java
class AutocompleteSystem {

	TrieNode root, curr;
	StringBuffer sb;

	public AutocompleteSystem(String[] sentences, int[] times) {
		if (sentences == null || times == null || sentences.length != times.length) {
			return;
		}

		reset();

		root = new TrieNode();

		for (int i = 0; i < times.length; i++) {
			insert(sentences[i], times[i]);
		}
	}

	public List<String> input(char c) {
		List<String> result = new ArrayList<>();
		if (curr == null) curr = root;
		if (c == '#') { // save sentence and reset state
			insert(sb.toString(), 1);
			reset();
			return result;
		}

		// Update global variable (curr TrieNode and string buffer); or append new character if not exist.
		sb.append(c);
		curr.children.putIfAbsent(c, new TrieNode());
		curr = curr.children.get(c);

		// MinHeap to find top 3.
		result.addAll(findTopK(curr, 3));

		return result;
	}

	private List<String> findTopK(TrieNode node, int k) {
		List<String> result = new ArrayList<>();
		if (node.freq.isEmpty()) {
			return result;
		}

		PriorityQueue<Pair> queue = new PriorityQueue<>(
			(a, b) -> a.count == b.count ? b.s.compareTo(a.s) : a.count - b.count);

		for (Map.Entry<String, Integer> entry : node.freq.entrySet()) {
			if (queue.size() < 3 || entry.getValue() >= queue.peek().count) {
				queue.offer(new Pair(entry.getKey(), entry.getValue()));
			}
			if (queue.size() > 3) queue.poll();
		}

		while (!queue.isEmpty()) {
			result.add(0, queue.poll().s);
		}

		return result;
	}

	private void reset() {
		curr = null;
		sb = new StringBuffer();
	}

	private void insert(String sentence, int count) {
		if (sentence == null || sentence.length() == 0) {
			return;
		}

		TrieNode node = root;
		for (char c : sentence.toCharArray()) {
			node.children.putIfAbsent(c, new TrieNode());
			node = node.children.get(c);
			node.freq.put(sentence, node.freq.getOrDefault(sentence, 0) + count);
		}
		node.isEnd = true; // can set word to node as well, if needed
	}
}

class TrieNode {
	public boolean isEnd;
	public Map<String, Integer> freq;
	public Map<Character, TrieNode> children; // Map is more applicable to all chars, not limited to 256 ASCII
	public TrieNode() {
		this.freq = new HashMap<>();
		this.children = new HashMap<>();
	}
}

class Pair {
	String s;
	int count;
	public Pair(String s, int count) {
		this.s = s;
		this.count = count;
	}
}
Python
class TrieNode:
    def __init__(self):
        self.is_end: bool = False
        self.freq: Dict[str, int] = {}
        self.children: Dict[str, TrieNode] = {}

class Pair:
    def __init__(self, sentence: str, freq: int):
        self.sentence: str = sentence
        self.freq: int = freq
        
    def __lt__(self, other):
        # Compare frequencies first, then lexicographical order
        if self.freq == other.freq:
            return self.sentence > other.sentence
        return self.freq < other.freq

class AutocompleteSystem:
    def __init__(self, sentences: List[str], times: List[int]) -> None:
        self.root = TrieNode()
        self.sb = ""  # string buffer to store the current sentence
        self.curr = None  # pointer to current TrieNode
        if sentences and times and len(sentences) == len(times):
            for i in range(len(sentences)):
                self._insert(sentences[i], times[i])

    def input(self, c: str) -> List[str]:
        result = []
        if not self.curr:
            self.curr = self.root

        if c == '#':  # save sentence and reset state
            self._insert(self.sb, 1)
            self._reset()
            return result

        # Update global variables to maintain state
        self.sb += c
        if c not in self.curr.children:
            self.curr.children[c] = TrieNode()
        self.curr = self.curr.children[c]
        
        # Find top K sentences using min-heap
        result = self._find_top_k(self.curr, 3)
        
        return result

    def _find_top_k(self, node: TrieNode, k: int) -> List[str]:
        result: List[str] = []
        if not node.freq:
            return result

        queue: List[Pair] = []
        for sentence, freq in node.freq.items():
            heapq.heappush(queue, Pair(sentence, freq))
            if len(queue) > k:
                heapq.heappop(queue)

        while queue:
            result.insert(0, heapq.heappop(queue).sentence)  # insert in reverse order
        
        return result

    def _reset(self) -> None:
        self.sb = ""
        self.curr = None
    
    def _insert(self, sentence: str, count: int) -> None:
        if not sentence:
            return
        
        node = self.root
        for char in sentence:
            if char not in node.children:
                node.children[char] = TrieNode()
            node = node.children[char]
            node.freq[sentence] = node.freq.get(sentence, 0) + count
        
        node.is_end = True

Time Complexity

  1. Insertion (_insert):

    • Each sentence of length m requires O(m) traversal of the Trie, with an additional update to the frequency map (search/update: O(log(k)) for k sentences). Hence, the complexity isO(m * log(k))

  2. Input Query (input):

    • Traversing prefix length p results in O(p).
    • Finding the top 3 sentences via heapq requires O(k * log(k)) for k sentences. Hence, the complexity is O(p * log(k)).

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