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 | int solution(const string& start, const string& target, |
| 2 | const vector<string>& words) { |
| 3 | |
| 4 | unordered_set<string> pool(words.begin(), words.end()); |
| 5 | |
| 6 | if (!pool.count(target)) { |
| 7 | return 0; |
| 8 | } |
| 9 | |
| 10 | queue<string> edge; |
| 11 | edge.push(start); |
| 12 | |
| 13 | int steps = 1; |
| 14 | |
| 15 | while (!edge.empty()) { |
| 16 | |
| 17 | int wide = edge.size(); |
| 18 | |
| 19 | for (int i = 0; i < wide; i++) { |
| 20 | |
| 21 | string word = edge.front(); |
| 22 | edge.pop(); |
| 23 | |
| 24 | if (word == target) { |
| 25 | return steps; |
| 26 | } |
| 27 | |
| 28 | for (int j = 0; j < (int)word.size(); j++) { |
| 29 | |
| 30 | char had = word[j]; |
| 31 | |
| 32 | for (char c = 'a'; c <= 'z'; c++) { |
| 33 | |
| 34 | word[j] = c; |
| 35 | |
| 36 | if (pool.count(word)) { |
| 37 | pool.erase(word); |
| 38 | edge.push(word); |
| 39 | } |
| 40 | } |
| 41 | |
| 42 | word[j] = had; |
| 43 | } |
| 44 | } |
| 45 | |
| 46 | steps++; |
| 47 | } |
| 48 | |
| 49 | return 0; |
| 50 | } |
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.