Formatted question description: https://leetcode.ca/all/1268.html
1268. Search Suggestions System (Medium)
Given an array of strings products
and a string searchWord
. We want to design a system that suggests at most three product names from products
after each character of searchWord
is typed. Suggested products should have common prefix with the searchWord. If there are more than three products with a common prefix return the three lexicographically minimums products.
Return list of lists of the suggested products
after each character of searchWord
is typed.
Example 1:
Input: products = ["mobile","mouse","moneypot","monitor","mousepad"], searchWord = "mouse" Output: [ ["mobile","moneypot","monitor"], ["mobile","moneypot","monitor"], ["mouse","mousepad"], ["mouse","mousepad"], ["mouse","mousepad"] ] Explanation: products sorted lexicographically = ["mobile","moneypot","monitor","mouse","mousepad"] After typing m and mo all products match and we show user ["mobile","moneypot","monitor"] After typing mou, mous and mouse the system suggests ["mouse","mousepad"]
Example 2:
Input: products = ["havana"], searchWord = "havana" Output: [["havana"],["havana"],["havana"],["havana"],["havana"],["havana"]]
Example 3:
Input: products = ["bags","baggage","banner","box","cloths"], searchWord = "bags" Output: [["baggage","bags","banner"],["baggage","bags","banner"],["baggage","bags"],["bags"]]
Example 4:
Input: products = ["havana"], searchWord = "tatiana" Output: [[],[],[],[],[],[],[]]
Constraints:
1 <= products.length <= 1000
- There are no repeated elements in
products
. 1 <= Σ products[i].length <= 2 * 10^4
- All characters of
products[i]
are lower-case English letters. 1 <= searchWord.length <= 1000
- All characters of
searchWord
are lower-case English letters.
Related Topics:
String
Solution 1. Trie
// OJ: https://leetcode.com/problems/search-suggestions-system/
// Time: O(DS) where D is the size of all the content in the `products`, S is the length of `searchWord`.
// Space: O(D)
struct TrieNode {
TrieNode *next[26] = {};
int index = -1;
};
class Solution {
void add(TrieNode *node, string &w, int i) {
for (char c : w) {
int j = c - 'a';
if (node->next[j] == NULL) node->next[j] = new TrieNode();
node = node->next[j];
}
node->index = i;
}
void collect(TrieNode *node, vector<string> &ans, vector<string> &A) {
if (ans.size() == 3) return;
if (!node) return;
if (node->index > -1) ans.push_back(A[node->index]);
for (int i = 0; i < 26; ++i) collect(node->next[i], ans, A);
}
public:
vector<vector<string>> suggestedProducts(vector<string>& A, string s) {
TrieNode root, *node = &root;
for (int i = 0; i < A.size(); ++i) add(&root, A[i], i);
vector<vector<string>> ans;
for (char c : s) {
ans.emplace_back();
if (ans.size() > 1 && ans[ans.size() - 2].empty()) continue;
if (node->next[c - 'a'] == NULL) continue;
node = node->next[c - 'a'];
collect(node, ans.back(), A);
}
return ans;
}
};
Java
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class Solution { public List<List<String>> suggestedProducts(String[] products, String searchWord) { List<List<String>> suggestedProducts = new ArrayList<List<String>>(); Arrays.sort(products); int productsCount = products.length; int searchLength = searchWord.length(); StringBuffer search = new StringBuffer(); for (int i = 0; i < searchLength; i++) { search.append(searchWord.charAt(i)); String curSearch = search.toString(); List<String> curSuggestedProducts = new ArrayList<String>(); int suggestCount = 0; for (int j = 0; j < productsCount; j++) { String product = products[j]; if (product.indexOf(curSearch) == 0) { curSuggestedProducts.add(product); suggestCount++; if (suggestCount == 3) break; } } suggestedProducts.add(curSuggestedProducts); } return suggestedProducts; } }
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Todo
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# 1268. Search Suggestions System # https://leetcode.com/problems/search-suggestions-system/ class Solution: def suggestedProducts(self, products: List[str], word: str) -> List[List[str]]: res = [] mp = collections.defaultdict(list) n = len(word) for i in range(n): w = word[:i+1] for p in products: if i < len(p) and p[:i+1] == w: mp[i].append(p) for key in range(n): results = sorted(mp[key]) tmp = [] for i in range(min(3, len(results))): tmp.append(results[i]) res.append(tmp) return res