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
searchWordare lower-case English letters.
class Solution { public List<List<String>> suggestedProducts(String[] products, String searchWord) { List<List<String>> results = new ArrayList<>(); if(products == null || products.length == 0) return results; // sort the products lexicographically Arrays.sort(products); for(int i = 1; i <= searchWord.length(); i++) { List<String> temp = new ArrayList<>(); int count = 0; for(String product : products) { if(product.length() >= i && product.substring(0, i).equals(searchWord.substring(0, i))) { temp.add(product); count++; if(count >= 3) { break; } } } results.add(temp); } return results; } }