Type an input and write what you think is the output . Each problem is converted to Web Assembly, so any possible input will show the corresponding output. Only correct predictions count.
Locked until you have predicted 3 outputs correctly.
| 1 | struct Trie { |
| 2 | Trie* kids[26] = {}; |
| 3 | bool word = false; |
| 4 | }; |
| 5 | |
| 6 | void insert(Trie* node, const string& text) { |
| 7 | |
| 8 | for (char c : text) { |
| 9 | |
| 10 | int i = c - 'a'; |
| 11 | |
| 12 | if (!node->kids[i]) { |
| 13 | node->kids[i] = new Trie(); |
| 14 | } |
| 15 | |
| 16 | node = node->kids[i]; |
| 17 | } |
| 18 | |
| 19 | node->word = true; |
| 20 | } |
| 21 | |
| 22 | Trie* step(Trie* node, const string& text) { |
| 23 | |
| 24 | for (char c : text) { |
| 25 | |
| 26 | int i = c - 'a'; |
| 27 | |
| 28 | if (!node->kids[i]) { |
| 29 | return nullptr; |
| 30 | } |
| 31 | |
| 32 | node = node->kids[i]; |
| 33 | } |
| 34 | |
| 35 | return node; |
| 36 | } |
| 37 | |
| 38 | vector<string> solution(const vector<string>& ops, |
| 39 | const vector<string>& args) { |
| 40 | |
| 41 | Trie root; |
| 42 | vector<string> ans; |
| 43 | |
| 44 | int n = ops.size(); |
| 45 | |
| 46 | for (int i = 0; i < n; i++) { |
| 47 | |
| 48 | if (ops[i] == "insert") { |
| 49 | insert(&root, args[i]); |
| 50 | ans.push_back("null"); |
| 51 | |
| 52 | } else if (ops[i] == "search") { |
| 53 | |
| 54 | Trie* at = step(&root, args[i]); |
| 55 | |
| 56 | if (at && at->word) { |
| 57 | ans.push_back("true"); |
| 58 | } else { |
| 59 | ans.push_back("false"); |
| 60 | } |
| 61 | |
| 62 | } else { |
| 63 | |
| 64 | Trie* at = step(&root, args[i]); |
| 65 | |
| 66 | if (at) { |
| 67 | ans.push_back("true"); |
| 68 | } else { |
| 69 | ans.push_back("false"); |
| 70 | } |
| 71 | } |
| 72 | } |
| 73 | |
| 74 | return ans; |
| 75 | } |
Names have been stripped. The signature is the only clue you get for free. Compiled as C++20 with the standard headers and using namespace std; already in scope.