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 | bool solution(const string& s, const string& p) { |
| 2 | |
| 3 | int m = s.size(); |
| 4 | int n = p.size(); |
| 5 | |
| 6 | vector<vector<bool>> ok(m + 1, vector<bool>(n + 1, false)); |
| 7 | ok[0][0] = true; |
| 8 | |
| 9 | for (int j = 2; j <= n; j++) { |
| 10 | |
| 11 | if (p[j - 1] == '*') { |
| 12 | ok[0][j] = ok[0][j - 2]; |
| 13 | } |
| 14 | } |
| 15 | |
| 16 | for (int i = 1; i <= m; i++) { |
| 17 | for (int j = 1; j <= n; j++) { |
| 18 | |
| 19 | if (p[j - 1] == '*') { |
| 20 | |
| 21 | if (j < 2) { |
| 22 | continue; |
| 23 | } |
| 24 | |
| 25 | bool skip = ok[i][j - 2]; |
| 26 | bool same = p[j - 2] == '.' || p[j - 2] == s[i - 1]; |
| 27 | |
| 28 | ok[i][j] = skip || (same && ok[i - 1][j]); |
| 29 | |
| 30 | } else if (p[j - 1] == '.' || p[j - 1] == s[i - 1]) { |
| 31 | ok[i][j] = ok[i - 1][j - 1]; |
| 32 | } |
| 33 | } |
| 34 | } |
| 35 | |
| 36 | return ok[m][n]; |
| 37 | } |
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.